-
Notifications
You must be signed in to change notification settings - Fork 0
/
form.py
1397 lines (1221 loc) · 69.2 KB
/
form.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
380
381
382
383
384
385
386
387
388
389
390
391
392
393
394
395
396
397
398
399
400
401
402
403
404
405
406
407
408
409
410
411
412
413
414
415
416
417
418
419
420
421
422
423
424
425
426
427
428
429
430
431
432
433
434
435
436
437
438
439
440
441
442
443
444
445
446
447
448
449
450
451
452
453
454
455
456
457
458
459
460
461
462
463
464
465
466
467
468
469
470
471
472
473
474
475
476
477
478
479
480
481
482
483
484
485
486
487
488
489
490
491
492
493
494
495
496
497
498
499
500
501
502
503
504
505
506
507
508
509
510
511
512
513
514
515
516
517
518
519
520
521
522
523
524
525
526
527
528
529
530
531
532
533
534
535
536
537
538
539
540
541
542
543
544
545
546
547
548
549
550
551
552
553
554
555
556
557
558
559
560
561
562
563
564
565
566
567
568
569
570
571
572
573
574
575
576
577
578
579
580
581
582
583
584
585
586
587
588
589
590
591
592
593
594
595
596
597
598
599
600
601
602
603
604
605
606
607
608
609
610
611
612
613
614
615
616
617
618
619
620
621
622
623
624
625
626
627
628
629
630
631
632
633
634
635
636
637
638
639
640
641
642
643
644
645
646
647
648
649
650
651
652
653
654
655
656
657
658
659
660
661
662
663
664
665
666
667
668
669
670
671
672
673
674
675
676
677
678
679
680
681
682
683
684
685
686
687
688
689
690
691
692
693
694
695
696
697
698
699
700
701
702
703
704
705
706
707
708
709
710
711
712
713
714
715
716
717
718
719
720
721
722
723
724
725
726
727
728
729
730
731
732
733
734
735
736
737
738
739
740
741
742
743
744
745
746
747
748
749
750
751
752
753
754
755
756
757
758
759
760
761
762
763
764
765
766
767
768
769
770
771
772
773
774
775
776
777
778
779
780
781
782
783
784
785
786
787
788
789
790
791
792
793
794
795
796
797
798
799
800
801
802
803
804
805
806
807
808
809
810
811
812
813
814
815
816
817
818
819
820
821
822
823
824
825
826
827
828
829
830
831
832
833
834
835
836
837
838
839
840
841
842
843
844
845
846
847
848
849
850
851
852
853
854
855
856
857
858
859
860
861
862
863
864
865
866
867
868
869
870
871
872
873
874
875
876
877
878
879
880
881
882
883
884
885
886
887
888
889
890
891
892
893
894
895
896
897
898
899
900
901
902
903
904
905
906
907
908
909
910
911
912
913
914
915
916
917
918
919
920
921
922
923
924
925
926
927
928
929
930
931
932
933
934
935
936
937
938
939
940
941
942
943
944
945
946
947
948
949
950
951
952
953
954
955
956
957
958
959
960
961
962
963
964
965
966
967
968
969
970
971
972
973
974
975
976
977
978
979
980
981
982
983
984
985
986
987
988
989
990
991
992
993
994
995
996
997
998
999
1000
# coding=utf-8
from __future__ import absolute_import
from __future__ import division
from __future__ import print_function
from __future__ import unicode_literals
from quantities import functions, known_units
from html import escape
selector = '''
<select %s onchange="insertAtCaret('commands',this.options[this.selectedIndex].value, %s);">
%s
</select>'''
selectoroption = '''
<option value="%s">%s</option>'''
def fill_selector(header, choices, backup=0):
html = ['<option value="" selected="selected" disable="disabled">%s</option>' % header]
for term in choices:
html.append(selectoroption % (term, term))
return selector % ("", backup, "".join(html))
# unit selectors, pre-baked
unit_list = list(known_units.keys())
unit_list.sort(key=lambda x: x.lower())
unit_selector = fill_selector("units", unit_list)
# function selectors, pre-baked
function_list = [f + "()" for f in functions]
function_list.extend(["using ", "in "])
function_selector = fill_selector("functions", function_list, backup=1)
# special symbol selector, pre-baked
symbol_list = ["⌗ve","μ", "π","ℎν ％ℳ","ΔᵣG°′","αβγδεζησχ","∞∡ℏ","äöüßø", "ΓΘΛΦΨ"]
symbol_selector = fill_selector("symbols", symbol_list)
def quant_selectors(known):
""" Make the selector for known quantities
"""
choices = known.split("\n")
return fill_selector("quantities", [term.rsplit("=",1)[0] for term in choices])
example_template = '''
<h3>How to use PQcalc</h3>
<p>PQcalc is an online calculator for students learning science in college.
To learn how to use it by example, click on any of the examples below, look at the
input, then press go and study the output. A more comprehensive documentation of
the program is <a href="http://www.bioinformatics.org/pqcalc/manual">here</a>.
<h3>Example calculations</h3><pre>%s</pre>
'''
def helpform(mob):
return example_template % exhtml
def newform(outp, logp, mem, known, log, mob, oneline, inputlog, prefill="", linespace="100%", logo=""):
mem = "\n".join(mem)
known = "\n".join(known)
if not outp:
logo = PQlogo
mem = mem.replace('"', '"')
inputlog = inputlog.replace('"', '"')
logbook = log.replace('"', '"') + "\n" + ("\n".join(logp)).replace('"', '"')
out = log.replace('"', '"') + "\n".join(outp)
keyb = "" if mob else 'class="keyboardInput"'
selectors = quant_selectors(known)
selectors = selectors + unit_selector + function_selector + symbol_selector
rows = 3
if prefill:
prefill = exdict[prefill][:-2]
rows =len(prefill.split("\n"))
data = dict(output=out, memory=mem, rows=rows, selectors=selectors, logbook=logbook, keyboard=keyb,
prefill=prefill, head=head, buttons=buttons, linespacing=linespace, logo=logo, inputlog=inputlog)
if oneline and not prefill:
return template_oneline % data
return template % data
def printableLog(symbols, logbook, inputlog):
known = "\n".join([s + "=" + si.__str__() for s, si in symbols.items()])
return printable_view % (known, escape(inputlog), logbook)
head = '''<head>
<meta name="format-detection" content="telephone=no">
<meta name="viewport" content="width=device-width; initial-scale=0.8;">
<link rel="shortcut icon" href="/ico/favicon.ico">
<script type="text/javascript"
src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML">
</script>
<script type="text/x-mathjax-config">
MathJax.Hub.Config({
TeX: { extensions: ["mhchem.js"] },
"HTML-CSS": {scale: 85}
});
</script>
<script type="text/javascript">
function goToAnchor() {
location.href = "#commands";
}
</script>
<script type="text/javascript">
<!--
function insertAtCaret(areaId,text,backup) {
var txtarea = document.getElementById(areaId);
var scrollPos = txtarea.scrollTop;
var strPos = 0;
var br = ((txtarea.selectionStart || txtarea.selectionStart == '0') ?
"ff" : (document.selection ? "ie" : false ) );
if (br == "ie") {
txtarea.focus();
var range = document.selection.createRange();
range.moveStart ('character', -txtarea.value.length);
strPos = range.text.length;
}
else if (br == "ff") strPos = txtarea.selectionStart;
var front = (txtarea.value).substring(0,strPos);
var back = (txtarea.value).substring(strPos,txtarea.value.length);
txtarea.value=front+text+back;
strPos = strPos + text.length - backup;
if (br == "ie") {
txtarea.focus();
var range = document.selection.createRange();
range.moveStart ('character', -txtarea.value.length);
range.moveStart ('character', strPos);
range.moveEnd ('character', 0);
range.select();
}
else if (br == "ff") {
txtarea.selectionStart = strPos;
txtarea.selectionEnd = strPos;
txtarea.focus();
}
txtarea.scrollTop = scrollPos;
}
</script>
<style>
@media all {
.page-break { display: none; }
}
@media print {
.page-break { display: block; page-break-before: always; }
}
</style>
</head>
'''
buttons = '''
<input type="button" value="+" onclick="insertAtCaret('commands','+',0);" />
<input type="button" value="-" onclick="insertAtCaret('commands','-',0);" />
<input type="button" value="*" onclick="insertAtCaret('commands',' * ',0);" />
<input type="button" value="/" onclick="insertAtCaret('commands',' / ',0);" />
<input type="button" value="^" onclick="insertAtCaret('commands','^',0);" />
<input type="button" value="()" onclick="insertAtCaret('commands','()', 1);" />
<input type="button" value="=" onclick="insertAtCaret('commands',' = ', 0);" />
<input type="button" value="[]" onclick="insertAtCaret('commands','[]', 1);" />
'''
rest ='''
'''
PQlogo = '''<table><tr><td>
<img src="data:image/gif;base64,R0lGODlh1wCSAPcAAAAAAAAAMwAAZgAAmQAAzAAA/wArAAArMwArZgArmQArzAAr/wBVAABVMwBVZgBVmQBVzABV/wCAAACAMwCAZgCAmQCAzACA/wCqAACqMwCqZgCqmQCqzACq/wDVAADVMwDVZgDVmQDVzADV/wD/AAD/MwD/ZgD/mQD/zAD//zMAADMAMzMAZjMAmTMAzDMA/zMrADMrMzMrZjMrmTMrzDMr/zNVADNVMzNVZjNVmTNVzDNV/zOAADOAMzOAZjOAmTOAzDOA/zOqADOqMzOqZjOqmTOqzDOq/zPVADPVMzPVZjPVmTPVzDPV/zP/ADP/MzP/ZjP/mTP/zDP//2YAAGYAM2YAZmYAmWYAzGYA/2YrAGYrM2YrZmYrmWYrzGYr/2ZVAGZVM2ZVZmZVmWZVzGZV/2aAAGaAM2aAZmaAmWaAzGaA/2aqAGaqM2aqZmaqmWaqzGaq/2bVAGbVM2bVZmbVmWbVzGbV/2b/AGb/M2b/Zmb/mWb/zGb//5kAAJkAM5kAZpkAmZkAzJkA/5krAJkrM5krZpkrmZkrzJkr/5lVAJlVM5lVZplVmZlVzJlV/5mAAJmAM5mAZpmAmZmAzJmA/5mqAJmqM5mqZpmqmZmqzJmq/5nVAJnVM5nVZpnVmZnVzJnV/5n/AJn/M5n/Zpn/mZn/zJn//8wAAMwAM8wAZswAmcwAzMwA/8wrAMwrM8wrZswrmcwrzMwr/8xVAMxVM8xVZsxVmcxVzMxV/8yAAMyAM8yAZsyAmcyAzMyA/8yqAMyqM8yqZsyqmcyqzMyq/8zVAMzVM8zVZszVmczVzMzV/8z/AMz/M8z/Zsz/mcz/zMz///8AAP8AM/8AZv8Amf8AzP8A//8rAP8rM/8rZv8rmf8rzP8r//9VAP9VM/9VZv9Vmf9VzP9V//+AAP+AM/+AZv+Amf+AzP+A//+qAP+qM/+qZv+qmf+qzP+q///VAP/VM//VZv/Vmf/VzP/V////AP//M///Zv//mf//zP///wAAAAAAAAAAAAAAACH5BAEAAPwALAAAAADXAJIAAAj/APcJHEiwoMGDCBMqXMiwocOHECNKnEixosWLGDNq3MixY0R9+PDduwfPnjF78JiVPJlyJUqVJl+6bBmTJkuYN2filFlzp02eOXvqHCq0aNCjQJP+XOqTWbx7IfV5VAgSnjGrV5sSRcp0q1KtRr967RqWLFewZ8eilZkVHj6pUwfmk8ksZL58+vLq3cu3r9+/gAMLHky4sOHDiAHfDRlPpcp8U/XFM2bsnrzEmDNr3sy5s2d9+fA1hncPrkZ8Ke3h/cy6tevXsPvma2b1nsaYl2Pr3s27t2DUlU1P1KfymO/jyJPHvndVOER98Egrn069emLm9pw7VOnMuvfv4PcC/9e+MJ708OjTK3dGGiLJZurjy+fNPJ5DfSfn69/POl9K8gY1lht/BBaIWD4qMZRPfgY26KBgjUGmEG2rPWjhhaBFp5B/8GHo4YMUJiSPPQN+aCJ/qDmT0GMntsjfVQjpY49xLtYoX4QHyWPMWzb2iB5q+BzEHo/yKaMJHEYAAQQEFjzApJJGaKIMND6q54w9yBzEXIXhQXMkBBU0KWYFTIr5gJgQAAGHMvEl80kddBAhpwYZaKCBnHTU8QmV8/lnm0EkcVkYHHCogWShRhCqJpKGRlmYMnAweaYFTJZZ5qSWmqkmm959QoeddIJa56h1hpqBnMnwiZ5/xxzEoGGgnP8pK5i0zkommXAMBgoQslpAaZi9nlkpsBaQ6euvFqw5XR1ElJrBqBp8YKoGIEBrrQZ0cCpYMku4QUcRdLjR7bfhjvtGHc9gdtJBViG2JKXHVuprpk3mCpgyRgg7L5r7ImusvGbC+wAQ2vL2ybSlSgutwnRWG6rDoSqRjGDQOGunxaHSWacbiSVokEqInUmsyGOCCW+ygEVq7K1iBvwvvCzr+2uZFdir28HPTstwxtBmzPC1dKjaVzIaI2x0qKUgZtVBJiEGL8DyWkrmAzbvpQyvJ2cKMJRvEFoEEPuaLHaxlRIM2zNFX0yttQ5L2/C1ozrs7Cd/VUzt221H+zaddB//tq5B7R6mCbG3mtzryVXnNfi8IxdbLBygDAapEZLOjOyZmrj2RqnR1rkztEQoQUSecSrRMM92/nynX5/APe3rGSR9WEquZocYKHBYAPbuuvdechF85T5s1skWDCvlx2L6a+KbmU6q2qbWIXtgz9ShhOewh9q3Xnb3nAEIOmecGEpMG/MavxBU/YPhAjdpR2fKrO+ryJYyn1gyRITq9sLoYlZK/glD2PT0gbbvYS9/pkvgnZSwPb/Bw1XweM27lqcXsPVrUvbTDCh+8LSZVQAIm0mG3NRmwDp0BhpEUB3EtldAtaXLNbQzCEpeQ7j05YWDTjLWmYAgNNbAoYYUxAza//RHRBOy5n97g1bfcGaqHnrmbwX5j2ssWKlcGSFrYYKAEXajjHg5jlIZDAw0SGgqIjjRM3sg46joNkQDamBirvFYQZrjmpLVDA69slTkeHM1fdGqAsYbTLMeVicQKGE3LRyhBp6hDO+dkTNLAxwzJIhFgIkJjr65YuHm9Ui/ABBhDYwNNK6XRA2UAnt1emFroEiQprmGWGUK2Jn2iJwJ/jGMfCGlG6PVydc4j4wI62VmSgJBSl6KX5SiZXKIta/MCeZgRgtlb375vexJszMmKR8N2QczMOGSi+070w8Eib2GXbM3ScxbKl8DI4O8ijVUZBy8vslHsjUzMKUwGh2sI/8q1K0Thsxgl+1aI0uYWQA8kSrZFuuWM7ktwTvP6BnDqiVMddlDm1Okl7Cc6R1oGEtgytxLHWAHnjq48Vrn3Az53BnBV7YMU8ALjyYCpkW/JMNi0jLid8I3qoompp1RnOQU59c+joKnX00KZOv0Nir05LOaFsMka2JYEFcSFFNZXCh6FqevkOqDmnQa4E7Z9s9VXpSlr0EeVumZHGjIS5x94RkIiKCep5JRrJ6RYyvN90qpUUo+HHwZX5YKLZ2iB5UX86nSjFE7SiJTPsJLnjJ/qTBVooeaDpPqZ1g5kOhs83JGRQ80gmUBo6JOA/JhYhNfQ1WCSLE1S8oipRSrnH5v1TQvBZSbYdNTypRqZqUFAVkdVwYm2iYHCIX7oF6Y6DnLpgezGTBuYYDqWr7CVmv6yV3W9HLKaQUyPHS4lmY909qBWJU1QNTPG9hXgeWerk76GanFnPvEs85xoPB8mnLnk9BKQYBPn7JWfONWJ7x2/0avnRXqVfel1fjEyp7tzYtJiagf5tJJuoSJZHBbCtuSjXM+M0UWBDjVLPDVia7zyW2oxtsZzgrEsy6VV4PV80OA6QWzh5yPhaPLWvu6VsHwZBKwQMjfYz5AL0uwFopTqzrfZiabH+Mwa+xZKfW2jEl6CbDaKkzIRbLzgYCzbpAvpx+wDYvI+lgqUwc8LRZDMqAyFPNnwvQyDPvGcZPSKs6gZefeBBhi9G2xjwfyzs+YGc+hRQ/AqMbdcmYg0OD5JdJeA9zqns+PFphxeDKBPlo2koS7RY/RIL0Z6hIav4b2VczkgzyyAXIvC1tdfCJKxAx8981Mk/Kco5bM+ORRd/+5RJ183vDeC7Nz0C+Ws2fA1isyxRQ9oDhWEN37MydPR3V1snbHwLzh88lWXvRIz/pY9l++tNCAOUYP0aDlMFJrRsM/Nqa/IPDs7ygjU2fyC1i9fNlil5U1Lt7Ha9Frsg6OODxZ/Jf9JlxGdavR2AA9iHA7XDmWfdg7oKAV487YvRWGR5dd1jZiKt1ZZXcGfWHzaluxyqQwgvUDS7YONBUGMR67xtQCKfSy5x2zW2eyckIWoz9Frpu0OatObt5MeQUyQ5d28OlJ3438uplovnzqc+7mzSCNRtFjs8vknEHuH12Wpj5rRmUBO+hgTAUxn+/mU0yt1giJbhgEC2TiU9b/mhebhGbfcJp4FnC7XhhOKhAsy3XTyjpm4J1gShbua38s0w8E/5ncCYtlbzCMLk2sgSGYPTGE9ffa3uj1MH/2WHCAxslkFvUTyk/IVTzMGL13J8V7ZsInxVYwe9zYoQqL0V18mbDYipmryVNSmh7MM+Q2d95I2ll1sBvD6F4YKFcV7JvRL8r0kXE6D89srXlDk8ZGpuQTZt1GC/Vn0I/YQ7awYbZHDM73QUfYkttmXTzmzIpAecL8PZYaF3ia8Qk1Z2IZIDGusWMRkxfds2KsBWdVpWue4UWMphcPBnv7QjX95xde4kEBSG+dMWE7EzGtVxhaBlUXk27rBmilF1Sv/4FDYpI4wRdO8VIzIqcMmZAvVxY2NfMZdvU8o/IG8cc6g3Qt0pJu+mA3pqJynEFyOYdqE5gpmdcXsbV6V8YkP0AomQAKygAKn3AkXfN0diQmTJgZKORoZEQHIpcMzCJRacgXP9h1N8dte+V4tJJBkfUyK2M5tDI27INpZbKBiGE6tcZ5EeMtn5CIz5CIcAJ3sRZrptQXDYh0D/h1p9dy91IExzQsi0Y89JOBrlZab+dPR2M0ZGWKGqAEgbaCoTKEhhFwOtcZWEMrUwgYu4JV7fNRMRM2OyiKvQENdDBCNOc6xHha0pJSk6gB1EcYdrcP5/UZyeMrbAUKOoiBQGdbu/8oJlXHGTcAAN5oU893MXlzWiioTuq3F+cmh63BeDkngbL4UTZUGNAABxxEM92UXJmyKbHhjd9YN0qAbWozjEwVTRX1fpR4c8jmjO7IGeFEfHsBDbhjZvDYL1kICp9HGPwIABTzCaazP9lzinRgYH8BCojlioWxdAIHZJ8RWPN0QsrQhVx4kZiRkYZRPaLzM4qkBAyEYedGJ2X4W3RIaAuZfZzokPxBk2b4DKbwCc8ADTKJfu32ZeyikssWgLVoIkgpWqpTgomBks/oGWEyM0a5H1kZHhb2ATLZF7A4lJrxLsYylvpRluChYtn2Gs34lSf3URV4InJpb47EWozlTlD/KIuWA5fz0ZcddXSkh5ACdXpnYpjygZjW8VScZ5IZlpB4GXZUdpWvgQliAAP8GANhgAmklglhEAMZeQNiEHWSqRemiZreKJqYEG6v0YDHSGmBOUdsmRkvtX2vMQwGkJHCCQDDwBfJIAnDKZxoABitCZzJyY+Z8BrsB3GrFJQvRpWE+TSQKRiY8JzCiUn5AJve6Y1h8BeSmQnjyY/z4BrJyJVKA4F12FfEsp2AMQypWZx5kQxowI+YhJ7kmQmYNAzdyI/4yReIaZ/8eAMFqp/8mYB604otSBD1N2aVwpmeEZzeKAZ/IQYauRcAEAaPFAb8WJ59gZgYCgAa6hf7+Rot16QwpkBp1umM2KcZ3hcm9OkX3RmbrJEMCeoXfemfABADyjGdNledvUdx83KjfcGh3lignoGYfSmiTaocBrmY6wifnbWbmCFLSmqg/OgaUPqlXtqPyWEKJRmhjTdU+9KleoGgAHADYCqmY9qhbdqjVPo6yzgYzYh3u1Y4bJqfdno/knAD4jmcPiqngOqNcDqkAclv65ibljZU8mKhm+GmkpAYmaAC6UmmHoqo+mCp04E2PyOIr4iZM5oZ3ieNrcGj3niph8Gkm0qnncqp+sCqAOCqyUGSiVeJ/4JpTMDyp58aqISRo4oKoHN6qLTqpouaHC1qJ3kqGE6YkucTNcBqqyRaGBmZoiXqqXkhl9Y6HUSzq3MocVqaGOMmJpS6Ga3ZF8rAjzDAnNyqD325rrHRk0U6VVj6YuXqNFkDrPJKoIWhrPBKq93KrRnppL5hpgNpmYMRcANnaCPjr8ipqPUJAHBkq8vKF0Aqq3rRlxP7phXrnjWJp3YZo3z6KKCwhZqQsqCwshmXNVHyhSyrCTKrsjL7CYJnqyiKo954sRkZSMRKsP9KsO3Kj9q6FzmKsJ0RrtjDsILBjgJ3quwqdpbjgUDUPm/Fiz/QSx/7ptGZF5ggngU6oN6ICf9tSrbxiphdewNfqw9h26Bi5CaJOLd0W7dvgjB6YrdzmxkBl5mAUY/21Iu8xjLy9C+wZAFqIBiw6p3Xqg9FO57iiawcqxeL+5yNyzrFSIwoiIpK1JUJGYuBUbX6Qj9AlLXy8m2uNhhdO5y4qhfJUKgZGQPHmbbxmherq5yCYa+cCzsjlLkimxfWF6mG0Zsi9nSwt0lUdjknQxjQMKjZ2rY4KqUAAANiYLa16o3v2hdSmr1147xGC72AoWYCKZDqNC0Ow3zTIpKAMX/3oBqHEVnDc7w05UU65GqZ0oNVEhihR47AhIbXwnZ8cxgL8icFcQzwICiDIX6f+IEXVDKAV3H/8Zi/gDF76fSID0pgPPU6xsUhB5EimAEN9EAlIuy4SRjCJCzCJjzCIKzCEiyP/2iKiuQ6buORDKOKiIEPJNLBO9LCHsLCLJwuPpyEBJSE+cDCmIEM8KAiBoEf+8rDTswZgYIQfvvEVNxi+ToQQFLFWswaIyIPCYEfpbHFYrwZIZIQJrGeY5zGAgy1BDEPzDCYahzHfXEMqsEQk1EicpzHejEXkPrFJ6vHcbwgEeQQIyK1gEzF5tEMEEEbYXzIaowMM/IRKoEMjpzG+AAyEgEdTlHJW2zAgzwcJXEMCMzJVUISzCAhFKEPdGwMo0zKLoLEwZER8YASrOzKNpIMpmwfwBsBDS1xDPjQyrZcIPkgD7mMyhzhHyYBD8cgD/iQDHcRzPpxF8ggGpRhEsY8FaFxEjiBFdx8FZRhFd0czt/szd4szuU8zuB8zum8zurczuz8zu4cz/A8z/Jcz/QMzzXxy3GBEMPsDE+BFDmxEwHtEgMdEzcREzHhGAmdGtrM0Pag0A39xhG90BL90A4N0RZd0Ri90Rfd0Rrt0RRN0S3RDPGAD/Jwzfuc0iq90izd0i790jAd0zI90zRd0zZ90w0REAA7"
width="115" height="76">
</td><td>
<h3> <br> 1. Set up a problem <br> 2. Solve it and press "go" <br> 3. See the work and the results </h3>
</td></tr></table><br>
'''
template = '''<html>
%(head)s
<body>
%(logo)s
<div style="line-height:%(linespacing)s">
%(output)s</div>
<div class="page-break"></div>
<form enctype="multipart/form-data" action="." method="post">
<textarea autofocus rows="%(rows)d" cols="42" id="commands" name="commands" %(keyboard)s type="number" autocapitalize="off"
autocomplete="on" spellcheck="false" style="font-weight: bold; font-size: 12pt;"
>%(prefill)s</textarea> <p>
<input type="submit" name="sub" style="font-weight: bold; font-size: 18pt;" value=" go ">
%(selectors)s </p>
%(buttons)s
<input type="submit" name = "sub" value="export" />
<input type="submit" name = "sub" value="reset" />
<input type="submit" name = "sub" value="help" />
<input type="hidden" name="memory" value = "%(memory)s"/>
<input type="hidden" name="inputlog" value = "%(inputlog)s"/>
<input type="hidden" name="logbook" value = "%(logbook)s"/>
</body>
</html>
'''
template_oneline = '''<html>
%(head)s
<body onload="goToAnchor();">
<table>
<tr><td><h1> PQCalc</h1></td><td>
<h3> : a scientific calculator that keeps track of units and uncertainties of physical quantities</h3>
</td></tr></table>
<table width=100%% cellspacing="2" cellpadding="10">
<tr>
<td width=100%% valign="top" style="border-width:6;border-color:#1132FF;border-style:ridge">
%(output)s
<form enctype="multipart/form-data" action="." method="post">
<input type="text" autofocus size="50" id="commands" name="commands" %(keyboard)s autocapitalize="off"
autocomplete="off" spellcheck="false" style="font-weight: bold; font-size: 12pt;">
<input type="submit" name="sub" value=" go ">
<p>
<a name="myAnchor" ></a>
%(buttons)s
<input type="submit" name = "sub" value="print" />
<input type="submit" name = "sub" value="reset" />
<input type="submit" name = "sub" value="help" />
</p>
%(selectors)s
<input type="hidden" name="memory" value = "%(memory)s"/>
<input type="hidden" name="inputlog" value = "%(inputlog)s"/>
<input type="hidden" name="logbook" value = "%(logbook)s"/>
</body>
</html>
'''
printable_view = '''<html>
<head>
<meta name="format-detection" content="telephone=no">
<script type="text/javascript"
src="http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS-MML_HTMLorMML">
</script>
<script type="text/x-mathjax-config">
MathJax.Hub.Config({ TeX: { extensions: ["mhchem.js"] }});
</script>
</head>
<body>Calculation done with PQCalc, the physical quantity calculator<br>
that knows about quantities, units and uncertainties.<br><br>
by Karsten Theis, Chemical and Physical Sciences, Westfield State University<br>
inquiries and bug reports to [email protected]<br><br>
Program currently hosted at
<a href="http://www.bioinformatics.org/pqcalc">
bioinformatics.org/PQCalc </a> and
<a href="http://ktheis.pythonanywhere.com/">
http://ktheis.pythonanywhere.com/ </a>
<h2> Known quantities </h2>
<PRE>%s</PRE>
<h2> Input </h2>
<PRE>%s</PRE>
<h2> Log of the calculation </h2>
%s
</div>
</body>
</html>
'''
example = '''problem = 1.71
# This has no units, no names like n[CH4], no sigfigs
0.6274*1.00e3/(2.205*2.54^3)
a_textbook = 17.4
problem = Units
t1 = 180 s
t2 = 4 min #notice the different unit
t_sum = t1 + t2
t_sum using s
days = min * 60 * 24
t_sum in days
problem = SigFigs Multiply
a_3sigfig = 2.63
a_5sigfig = 2.6300
b_4sigfig = 0.3128
c_less = a_3sigfig * b_4sigfig
c_more = a_5sigfig * b_4sigfig
problem = SigFigs Add
a_3sigfig = 2.63
a_5sigfig = 2.6300
b_4sigfig = 0.3128
c_less = a_3sigfig + b_4sigfig
c_more = a_5sigfig + b_4sigfig
problem = Addition
a = 5.1 s
b = 3.4 s
c = a + b
problem = Order of operations
a = -3 kJ / 5 mM * (4 atm + 5.4 Pa)
problem = Different numbers
a = (1/2) 1/s^2
__showuncert__ = 1
b = 2.341(4)
c = sqrt(a)
d = a^(1/3)
problem = Functions
a = 0.34
b = minimum(log(a), ln(a), exp(a), sin(a), cos(a), tan(a)) - maximum(absolute(-45.6(4)), quadp(1,2,-3))
problem = division
# 12.05 mol of dry ice have a mass of 530.4 g.
# What is the molar mass of [CO2]?
m[CO2] = 530.4 g
n[CO2]=12.05 mol
M[CO2] = m[CO2] / n[CO2]
problem = crash1
m = 0.0
b = .6 / m_
problem = 4.111
V_Analyte = 100.0 mL
V_Titrant = 3.19 mL
c_Titrant = 0.0250 M
# 1:1 stochiometry...
c_Analyte = c_Titrant * V_Titrant / V_Analyte
#answer: c_Analyte =7.98e-4 M
problem = GenChem constants
R = 8.314 J/(mol K)
T_room = 293.15 K
P_atm = 1 atm
q_e = 1.602176E-19 C
c0 = 299792458 m/s
F = 96485.33 C/mol
M_H = 1.08 g/mol
M_C = 12.011 g/mol
M_O = 15.999 g/mol
M_Na = 22.9897 g/mol
M_Cl = 35.45g/mol
M_N = 14.007 g/mol
N_A = 6.02214E23 /mol
problem = languages
# Chemical equation
# 化学方程式
# Химическое уравнение
# 化学反応式
# Reaktionsgleichung
# 화학반응식
# Ecuación química
# إنشاء حسابد خول
# Równanie reakcji
# Phương trình hóa học
! O2(g) + 2H2(g) -> 2H2O(g)
problem = 1.87
percent = 1/100
year = 365 * 24 h
input_Fertilizer = 1500. kg / year
input_N = 10 percent * input_Fertilizer
wash_N = 15 percent * input_N
flow_stream = 1.4 m^3/min
[N]added = wash_N / flow_stream
[N]added using mg L
a_textbook = 0.031 mg/L
problem = 1.96
pi = 3.14159265
d_gasoline = 0.73 g/mL
d_water = 1.00 g/mL
diameter = 3.2 cm
radius = diameter / 2
m_water = 34.0 g
m_gasoline = 34.0 g
V_water = m_water / d_water
V_gasoline = m_gasoline / d_gasoline
V_total = V_water + V_gasoline
h_total = V_total / (radius^2 * pi)
#######
problem = 3.91
#given
m[CxHy] = 135.0 mg
m[CO2] = 440.0 mg
m[H2O] = 135.0 mg
M[CxHy] = 270.0 g/mol
#calculated from chemical formula
M[C] = 12.0107 g/mol
M[O] = 15.9994 g/mol
M[H] = 1.00794 g/mol
M[CO2] = M[C] + 2 * M[O]
M[H2O] = 2 * M[H] + M[O]
#<div class="page-break"></div>
#calculation
n[CO2] = m[CO2] / M[CO2]
n[C] = n[CO2]
n[H2O] = m[H2O] / M[H2O]
n[H] = n[H2O]*2
n[CxHy] = m[CxHy] / M[CxHy]
ratio[nuC:nuH] = n[C] / n[H]
# _M[CxHy] = _M[C] * _nuC + _M[H] * _nuH = _M[C] * _ratio[nuC:nuH] _nuH + _M[H] * _nuH
# _M[CxHy] = _nuH * (_M[C] * _ratio[nuC:nuH] + _M[H] )
nuH = M[CxHy] / (M[C] * ratio[nuC:nuH] + M[H])
nuC = ratio[nuC:nuH] * nuH
#texbook answer: [C20H30]
######
problem = 3.103
# If 2.50 g of [KO2] react with
# 4.50 g of [CO2], how much dioxygen
# will form (together with [K2CO3])?
m[KO2] = 2.50 g
m[CO2] = 4.50 g
! 4KO2 + 2CO2 -> 3O2 + 2K2CO3
ν[KO2] = 4
ν[CO2] = 2
ν[O2] = 3
#calculate M from chemical formula
__skip__=1
M[C] = 12.0107 g/mol
M[O] = 15.9994 g/mol
M[K] = 39.0983 g/mol
M[KO2] = M[K] + 2 * M[O]
M[CO2] = M[C] + 2 * M[O]
#calculation
__skip__=0
n[KO2] = m[KO2] / M[KO2]
n[CO2] = m[CO2] / M[CO2]
n[KO2 -> lim] = n[KO2] / ν[KO2]
n[CO2 -> lim] = n[CO2] / ν[CO2]
n[lim] = minimum(n[KO2 -> lim], n[CO2 -> lim])
V[O2] = n[lim] ν[O2] 8.314 J/(K mol) 298 K / 1 atm
V[O2] using L
m[O2] = n[lim] ν[O2] * 2 * M[O]
# textbook : _m[O2] = 0.844 g
problem = 6.87
# What is the density of radon at 298 K and 1 atm?
T = 298 K
P = 1.00 atm
R = 8.314 J / (mol K)
R using atm L
M_Rn = 222. g / mol
# Consider _n_Rn = 1 mol
# Any other value would give the same result...
n_Rn = 1 mol
V0 = n_Rn * R T / P
m0 = n_Rn M_Rn
ρ = m0 / V0
problem = 17.27
[H+] = 3.45e-8 M
pH = -log([H+]/1 M)
#answer =
######
problem = 17.49
K_a = 1.80e-4
c = 0.0600 M
# K_a = x^2/(0.06-x) <=> x^2 + x K_a - 0.06 K_a c/M = 0
# solve quadratic: either quadp or quadn gives physically sensible answer
A_ = 1
B_ = K_a
C_ = - K_a c / 1 M
xp = quadp(A_, B_, C_)
xn = quadn(A_, B_, C_)
pH = -log(xn)
# using the approximation that c[AH] = c[total]
pH = -1/2 log(c / M K_a)
######
problem = 16.115
K1 = 4.10e-4
T1 = CtoKscale(2000.)
T2 = CtoKscale(25.)
ΔH = 180.6 kJ/mol
R = 8.314 J/(mol K)
# ln(_K2 / _K1) = -ΔH/R (1/_T2 - 1/_T1)
K2 = K1 exp(-ΔH/R (1/T2 - 1/T1))
######
problem = made.up
T = 298.0 K
R = 8.314 J/(mol K)
K_eq = 2.54e6
ΔG° = - R T ln(K_eq)
problem = first semester
5 + 3
__tutor__ = 1
6 + 8
R_gas = 0.08205746(14) L atm K^-1 mol^-1
R_ryd = 1.0973731568539(55)e7 m^-1
problem = practice units
__hideunits__ = 1
R = 8.3144621(75) J/(K mol)
T = 298.0 K
K_eq = 2.54e6
ΔG° = - R T ln(K_eq)
problem = second semester
R = 8.3144621(75) J/(K mol)
T = 298.0 K
K_eq = 2.54e6
ΔG° = - R T ln(K_eq)
problem = physical chemistry
__showuncert__= 1
# gas constant
R = 8.3144621(75) J/(K mol)
# Boltzmann constant
k_B = 1.3806488(13)e-23 J/K
# Plank constant
h = 6.62606957(29)e-34 J s
# Speed of light (nowadays not a measurement, but a definition)
c0 = 299792458 m/s
# Mathematical constant Pi to 17 digits
π = 3.1415926535897932
# Permeability in vacuum
μ0 = 4 N A^-2 π / 10000000
# Permittivity in vacuum
ε0 = 1 / (c0^2 μ0)
# mass of electron
m_elec= 9.10938215(45)e-31kg
# Elementary charge
e = 1.60217657e-19 C
# Rydberg constant from first principles
R_H = m_elec e^4 / (8 h_^3 ε0^2 c0)
#Bohr radius to check
ħ = h_ / (2π)
a0 = 4 π ε0 ħ^2 /(m_elec e^2)
problem = R_units
R = 8.314 Pa m^3/(mol K)
R2 = R / (101325 Pa * 1 m^3) * 1 atm * 1000 L
R3 = R * 1 atm * 1000 L / (101325 Pa * 1 m^3)
R4 = R * (1 atm / 101325 Pa) * (1000 L / 1 m^3)
P = 1 atm
P using Pa
problem = concentration
# Titration of KOH with [H2SO4] to equivalence point
# Given:
V[KOH] = 5.00 mL
[H2SO4] = 0.100 M
V[H2SO4] = 34.5 mL
# Plan: 1) _n[H2SO4] via c=n/V,
# 2) _n[KOH] via coefficients,
# 3) _[KOH] via c=n/V
n[H2SO4] = [H2SO4] * V[H2SO4]
!H2SO4 + 2 KOH -> 2 H2O + 2 K+ + 2 SO4^2-
n[KOH] = n[H2SO4] / 1 * 2
[KOH]original = n[KOH] / V[KOH]
problem = showcase chemistry markup
# Limiting reagent problem
# The balanced reaction is:
! H2SO4 + 2 KOH -> 2 H2O + 2 K+ + 2 SO4^2-
# Given:
n[H2SO4]start = 3.4 mmol
n[KOH]start = 7.4 mmol
n[->] = minimum(n[H2SO4]start/1, n[KOH]start/2)
n[H2SO4]end = n[H2SO4]start - 1 * n[->]
n[KOH]end = n[KOH]start - 2 * n[->]
problem = J Chem Ed 2012, 89, 326−334
# 0.564 grams of [AgNO3] is dissolved in 25.00 mL of 0.250 molar [BaCl2].
# A precipitate forms and is isolated and weighed.
# Its mass is 0.392 grams.
# What is the percent yield of the reaction?
m[AgNO3]= 0.564 g
V[BaCl2 sol] = 25.00 mL
c[BaCl2] = 0.250 M
m_prec= 0.392 g
# Plan:
# 1) Nature of precipitate
# 2) Find limiting reagent
# 3) Calculate yield!
#
# 1) Nature of precipitate
! BaCl2(s) -> Ba^2+(aq) + 2 Cl- (aq)
! AgNO3(s) -> Ag+(aq) + NO3- (aq)
# [AgNO3], [BaCl2], and [Ba(NO3)2] are soluble, [AgCl] isn't
#
# 2) Find limiting reagent
# First, need a balanced chemical equation
! 2AgNO3 + BaCl2 -> 2AgCl(s) + Ba^2+ + 2 NO3-
# Try either reagent in excess and see which gives least product
# Case A: [AgNO3] limiting and [BaCl2] in excess
M[AgNO3] = 169.87 g/mol
n[AgNO3] = m[AgNO3] / M[AgNO3]
# [AgNO3] and [AgCl] at equimolar ratio
n[AgCl]A = n[AgNO3]
# Case B: [AgNO3] in excess and [BaCl2] limiting
n[BaCl2] = c[BaCl2] V[BaCl2 sol]
# [BaCl2] and [AgCl] at 1:2 molar ratio
n[AgCl]B = n[BaCl2] * 2
# _n[AgCl]A is smaller than _n[AgCl]A , so [AgNO3] is limiting
#
# 3) Calculate yield
M[AgCl] = 143.32 g/mol
m[AgCl]theoretical = n[AgCl]A * M[AgCl]
m[AgCl]actual = m_prec
yield = m[AgCl]actual / m[AgCl]theoretical
problem = units with non-integer powers
# Inspired by http://www.chem.ox.ac.uk/vrchemistry/rates/newhtml/order3.htm
#The reaction of chlorine with chloroform to yield carbon tetrachloride and hydrogen chloride is
! Cl2 + CHCl3 -> CCl4 + HCl
# The observed rate expression for production of HCl is first order in chloroform and half order in chlorine. If the rate is 0.183 M/s and the concentrations of chloroform and chlorine are 0.400 M and 1.52 mM, respectively, what is the rate constant of the reaction?
rate = 0.183 M/s
[Cl2] = 1.52 mM
[CHCl3] = 0.400 M
k[->] = rate / ( [Cl2]^(1/2) * [CHCl3] )
problem = error in adding
a = 5.6 km + 25.1 mol
problem = error in power
a = 5.6 ^ (25 mol)
problem = error in division
a = 5.6 / 0.0
problem = error with parentheses
a = (5 + 6))
problem = error in minimum function
a = minimum(7 M, 34 mol)
problem = limitations: numeric overflow
a = 10^10^10
problem = warning in square root
#This works
a = sqrt(2.3 mm * 4.9 km)
#This works too, but results in a warning
a = sqrt(2.3 mm * 4.13 g)
problem = limitations: imaginary numbers
#PQcalc works with real numbers, only
a = sqrt(-1)
problem = error in log function
a = log(0.34 mol/L)
problem = error in log function
a = log(-23.8)
problem = tutorial
# You are asked to make about 80 mL of a 0.150 mol/L aqueous solution of magnesium chloride by mixing the pure solid with water.
V_requested = 80 mL
[MgCl2] = 0.150 mol/L
# First, calculate how much [MgCl2] you need.
n[MgCl2] = [MgCl2] V_requested
# Because we want to use a balance to measure the pure solid, we need to calculate the mass from the chemical amount.
M[MgCl2] = 95.211 g/mol
m[MgCl2]exact = n[MgCl2] M[MgCl2]
# You took a bit more than the required mass and placed 1.213 g of [MgCl2] into a 100 mL graduated cylinder.
#
# To which volume do you have to add water to obtain a 0.150 M solution?
V_exact = m[MgCl2]exact / (M[MgCl2] [MgCl2] )
m[MgCl2]actual = 1.213g
V_actual = m[MgCl2]actual / (M[MgCl2] [MgCl2])
# Another, simpler way of calculating: Instead of 1.143 g, you have 1.213 g, so you have to add more water to obtain a proportionally larger volume
V_made_2 = V_exact * m[MgCl2]actual / m[MgCl2]exact
problem = integrate images
#How does water form from the elements, and what is the chemical equation?
#{"http://bit.ly/1PvpnpD" width ="160"}
!H2 + O2 -> 2H2O
problem = 1.1* Calculation of Density
<h3>Example 1.1: Calculation of Density</h3>
#Gold — in bricks, bars, and coins — has been a form of currency for centuries. In order to swindle people into paying for a brick of gold without actually investing in a brick of gold, people have considered filling the centers of hollow gold bricks with lead to fool buyers into thinking that the entire brick is gold. It does not work: Lead is a dense substance, but its density is not as great as that of gold, 19.3 g/cm3. What is the density of lead if a cube of lead has an edge length of 2.00 cm and a mass of 90.7 g?
m_cube = 90.7 g
l_cube = 2.00 cm
d_gold = 19.3 g/cm^3
#<h4>Plan</h4>The density of an object can be calculated by dividing its mass by its volume. The volume of a cube is calculated by cubing the edge length. First, we will calculate the volume of the cube. Then, we will calculate the density of the lead cube.
V_cube = l_cube ^ 3
d_lead = m_cube / V_cube
#(We will discuss the reason for rounding to the first decimal place in the next section.)
#<h4>Check Your Learning</h4>
(a) To three decimal places, what is the volume of a cube (cm3) with an edge length of 0.843 cm?
(b) If the cube in part (a) is copper and has a mass of 5.34 g, what is the density of copper to two decimal places?
#Answer:
#(a) 0.599 cm^3; (b) 8.91 g/cm^3
problem = 1.2 Using Displacement of Water to Determine Density
<h3>Example 1.2: Using Displacement of Water to Determine Density</h3>
Using Displacement of Water to Determine Density
#This (http://openstaxcollege.org/l/16phetmasvolden)PhET simulation illustrates another way to determine density, using displacement of water. Determine the density of the red and yellow blocks.
#<h4>Plan</h4>When you open the density simulation and select Same Mass, you can choose from several 5.00-kg colored blocks that you can drop into a tank containing 100.00 L water. The yellow block floats (it is less dense than water), and the water level rises to 105.00 L. While floating, the yellow block displaces 5.00 L water, an amount equal to the weight of the block. The red block sinks (it is more dense than water, which has density = 1.00 kg/L), and the water level rises to 101.25 L.
#The red block therefore displaces 1.25 L water, an amount equal to the volume of the block. The density of the red block is:
#density=massvolume=5.00 kg1.25 L=4.00 kg/L
#Note that since the yellow block is not completely submerged, you cannot determine its density from this information. But if you hold the yellow block on the bottom of the tank, the water level rises to 110.00 L, which means that it now displaces 10.00 L water, and its density can be found:
#density=massvolume=5.00 kg10.00 L=0.500 kg/L<h4>Check Your Learning</h4>Remove all of the blocks from the water and add the green block to the tank of water, placing it approximately in the middle of the tank. Determine the density of the green block.
#Answer:
#2.00 kg/L
#<hr>
problem = 1.3 Rounding Numbers
<h3>Example 1.3: Rounding Numbers</h3>
Rounding Numbers
#Round the following to the indicated number of significant figures:
#(a) 31.57 (to two significant figures)
#(b) 8.1649 (to three significant figures)
#(c) 0.051065 (to four significant figures)
#(d) 0.90275 (to four significant figures)
#<h4>Plan</h4>(a) 31.57 rounds “up” to 32 (the dropped digit is 5, and the retained digit is even)
#(b) 8.1649 rounds “down” to 8.16 (the dropped digit, 4, is less than 5)(c) 0.051065 rounds “down” to 0.05106 (the dropped digit is 5, and the retained digit is even)
#(d) 0.90275 rounds “up” to 0.9028 (the dropped digit is 5, and the retained digit is even)
#<h4>Check Your Learning</h4>Round the following to the indicated number of significant figures:
#(a) 0.424 (to two significant figures)
#(b) 0.0038661 (to three significant figures)
#(c) 421.25 (to four significant figures)
#(d) 28,683.5 (to five significant figures)
#
#Answer:
#(a) 0.42; (b) 0.00387; (c) 421.2; (d) 28,684
#<hr>
problem = 1.4 Addition and Subtraction with Significant Figures
<h3>Example 1.4: Addition and Subtraction with Significant Figures</h3>
Addition and Subtraction with Significant Figures
#Rule: When we add or subtract numbers, we should round the result to the same number of decimal places as the number with the least number of decimal places (i.e., the least precise value in terms of addition and subtraction).
#(a) Add 1.0023 g and 4.383 g.
#(b) Subtract 421.23 g from 486 g.
#<h4>Plan</h4>
#(a) 1.0023 g+ 4.383 g5.3853 g
#Answer is 5.385 g (round to the thousandths place; three decimal places)
#(b) 486 g−421.23 g64.77 g
#Answer is 65 g (round to the ones place; no decimal places)
#
#
#
#<h4>Check Your Learning</h4>(a) Add 2.334 mL and 0.31 mL.
#(b) Subtract 55.8752 m from 56.533 m.
#
#Answer:
#(a) 2.64 mL; (b) 0.658 m
#<hr>
problem = 1.5 Multiplication and Division with Significant Figures
<h3>Example 1.5: Multiplication and Division with Significant Figures</h3>
Multiplication and Division with Significant Figures
#Rule: When we multiply or divide numbers, we should round the result to the same number of digits as the number with the least number of significant figures (the least precise value in terms of multiplication and division).
#(a) Multiply 0.6238 cm by 6.6 cm.
#(b) Divide 421.23 g by 486 mL.
#<h4>Plan</h4>
#(a) 0.6238 cm×6.6cm=4.11708cm2⟶result is4.1cm2(round to two significant figures)four significant figures×two significant figures⟶two significant figures answer
#(b) 421.23 g486 mL=0.86728... g/mL⟶result is 0.867 g/mL(round to three significant figures)five significant figuresthree significant figures⟶three significant figures answer
#<h4>Check Your Learning</h4>(a) Multiply 2.334 cm and 0.320 cm.
#(b) Divide 55.8752 m by 56.53 s.
#
#Answer:
#(a) 0.747 cm2 (b) 0.9884 m/s
#<hr>
problem = 1.6 Calculation with Significant Figures
<h3>Example 1.6: Calculation with Significant Figures</h3>
Calculation with Significant Figures
#One common bathtub is 13.44 dm long, 5.920 dm wide, and 2.54 dm deep. Assume that the tub is rectangular and calculate its approximate volume in liters.
#<h4>Plan</h4>
#V=l×w×d=13.44 dm×5.920 dm×2.54 dm=202.09459...dm3(value from calculator)=202 dm3, or 202 L(answer rounded to three significant figures)<h4>Check Your Learning</h4>What is the density of a liquid with a mass of 31.1415 g and a volume of 30.13 cm3?
#
#Answer:
#1.034 g/mL
#<hr>
problem = 1.7 Experimental Determination of Density Using Water Displacement
<h3>Example 1.7: Experimental Determination of Density Using Water Displacement</h3>
Experimental Determination of Density Using Water Displacement
#A piece of rebar is weighed and then submerged in a graduated cylinder partially filled with water, with results as shown.
#
#
#
#(a) Use these values to determine the density of this piece of rebar.
#(b) Rebar is mostly iron. Does your result in (a) support this statement? How?
#
#SolutionThe volume of the piece of rebar is equal to the volume of the water displaced:
#volume=22.4 mL−13.5 mL=8.9 mL=8.9 cm3(rounded to the nearest 0.1 mL, per the rule for addition and subtraction)
#The density is the mass-to-volume ratio:
#density=massvolume=69.658 g8.9 cm3=7.8 g/cm3
#(rounded to two significant figures, per the rule for multiplication and division)
#From Table 1.4, the density of iron is 7.9 g/cm3, very close to that of rebar, which lends some support to the fact that rebar is mostly iron.
#<h4>Check Your Learning</h4>An irregularly shaped piece of a shiny yellowish material is weighed and then submerged in a graduated cylinder, with results as shown.
#
#
#
#(a) Use these values to determine the density of this material.
#(b) Do you have any reasonable guesses as to the identity of this material? Explain your reasoning.
#
#Answer:
#(a) 19 g/cm3; (b) It is likely gold; the right appearance for gold and very close to the density given for gold in Table 1.4.
#<hr>
problem = 1.8 Using a Unit Conversion Factor
<h3>Example 1.8: Using a Unit Conversion Factor</h3>
Using a Unit Conversion Factor
#The mass of a competition frisbee is 125 g. Convert its mass to ounces using the unit conversion factor derived from the relationship 1 oz = 28.349 g (Table 1.6).
#<h4>Plan</h4>If we have the conversion factor, we can determine the mass in kilograms using an equation similar the one used for converting length from inches to centimeters.
#xoz=125 g×unit conversion factor
#We write the unit conversion factor in its two forms:
#1 oz28.349 gand28.349 g1 ozThe correct unit conversion factor is the ratio that cancels the units of grams and leaves ounces.
#xoz=125g×1 oz28.349g=(12528.349)oz=4.41 oz (three significant figures)
#<h4>Check Your Learning</h4>Convert a volume of 9.345 qt to liters.
#
#Answer:
#8.844 L
#<hr>
problem = 1.9 Computing Quantities from Measurement Results and Known Mathematical Relations
<h3>Example 1.9: Computing Quantities from Measurement Results and Known Mathematical Relations</h3>
Computing Quantities from Measurement Results and Known Mathematical Relations
#What is the density of common antifreeze in units of g/mL? A 4.00-qt sample of the antifreeze weighs 9.26 lb.
#<h4>Plan</h4>Since density=massvolume, we need to divide the mass in grams by the volume in milliliters. In general: the number of units of B = the number of units of A × unit conversion factor. The necessary conversion factors are given in Table 1.6: 1 lb = 453.59 g; 1 L = 1.0567 qt; 1 L = 1,000 mL. We can convert mass from pounds to grams in one step:
#9.26lb×453.59 g1lb=4.20×103g
#We need to use two steps to convert volume from quarts to milliliters.
#
#
#Convert quarts to liters.
#4.00qt×1 L1.0567qt=3.78 L
#
#Convert liters to milliliters.
#3.78L×1000 mL1L=3.78×103mL
#
#Then,
#density=4.20×103g3.78×103mL=1.11 g/mLAlternatively, the calculation could be set up in a way that uses three unit conversion factors sequentially as follows:
#9.26lb4.00qt×453.59 g1lb×1.0567qt1L×1L1000 mL=1.11 g/mL
#<h4>Check Your Learning</h4>What is the volume in liters of 1.000 oz, given that 1 L = 1.0567 qt and 1 qt = 32 oz (exactly)?
#
#Answer:
#2.956×10−2L
#
#<hr>
problem = 1.10 Computing Quantities from Measurement Results and Known Mathematical Relations
<h3>Example 1.10: Computing Quantities from Measurement Results and Known Mathematical Relations</h3>
Computing Quantities from Measurement Results and Known Mathematical Relations
#While being driven from Philadelphia to Atlanta, a distance of about 1250 km, a 2014 Lamborghini Aventador Roadster uses 213 L gasoline.
#(a) What (average) fuel economy, in miles per gallon, did the Roadster get during this trip?
#(b) If gasoline costs $3.80 per gallon, what was the fuel cost for this trip?
#<h4>Plan</h4>(a) We first convert distance from kilometers to miles:
#1250 km×0.62137 mi1 km=777 mi
#and then convert volume from liters to gallons:
#213L×1.0567qt1L×1 gal4qt=56.3 gal
#Then,
#
#(average) mileage=777 mi56.3 gal=13.8 miles/gallon=13.8 mpg
#Alternatively, the calculation could be set up in a way that uses all the conversion factors sequentially, as follows:
#1250km213L×0.62137 mi1km×1L1.0567qt×4qt1 gal=13.8 mpg
#(b) Using the previously calculated volume in gallons, we find:
#56.3 gal×$3.801 gal=$214
#<h4>Check Your Learning</h4>A Toyota Prius Hybrid uses 59.7 L gasoline to drive from San Francisco to Seattle, a distance of 1300 km (two significant digits).
#(a) What (average) fuel economy, in miles per gallon, did the Prius get during this trip?
#(b) If gasoline costs $3.90 per gallon, what was the fuel cost for this trip?
#
#Answer:
#(a) 51 mpg; (b) $62
#<hr>
problem = 1.11 Conversion from Celsius
<h3>Example 1.11: Conversion from Celsius</h3>
Conversion from Celsius
#Normal body temperature has been commonly accepted as 37.0 °C (although it varies depending on time of day and method of measurement, as well as among individuals). What is this temperature on the kelvin scale and on the Fahrenheit scale?
#<h4>Plan</h4>
#K=°C+273.15=37.0+273.2=310.2 K
#°F=95°C+32.0=(95×37.0)+32.0=66.6+32.0=98.6 °F
#<h4>Check Your Learning</h4>Convert 80.92 °C to K and °F.
#
#Answer:
#354.07 K, 177.7 °F
#<hr>
problem = 1.12 Conversion from Fahrenheit
<h3>Example 1.12: Conversion from Fahrenheit</h3>
Conversion from Fahrenheit
#Baking a ready-made pizza calls for an oven temperature of 450 °F. If you are in Europe, and your oven thermometer uses the Celsius scale, what is the setting? What is the kelvin temperature?
#<h4>Plan</h4>
#°C=59(°F−32)=59(450−32)=59×418=232 °C⟶set oven to 230 °C(two significant figures)
#K=°C+273.15=230+273=503 K⟶5.0×102K(two significant figures)
#<h4>Check Your Learning</h4>Convert 50 °F to °C and K.
#
#Answer:
#10 °C, 280 K
#<hr>
problem = 2.1 Testing Dalton’s Atomic Theory
<h3>Example 2.1: Testing Dalton’s Atomic Theory</h3>
Testing Dalton’s Atomic Theory
#In the following drawing, the green spheres represent atoms of a certain element. The purple spheres represent atoms of another element. If the spheres touch, they are part of a single unit of a compound. Does the following chemical change represented by these symbols violate any of the ideas of Dalton’s atomic theory? If so, which one?
#
#
#
#<h4>Plan</h4>The starting materials consist of two green spheres and two purple spheres. The products consist of only one green sphere and one purple sphere. This violates Dalton’s postulate that atoms are neither created nor destroyed during a chemical change, but are merely redistributed. (In this case, atoms appear to have been destroyed.)
#<h4>Check Your Learning</h4>In the following drawing, the green spheres represent atoms of a certain element. The purple spheres represent atoms of another element. If the spheres touch, they are part of a single unit of a compound. Does the following chemical change represented by these symbols violate any of the ideas of Dalton’s atomic theory? If so, which one?
#
#
#
#
#Answer:
#The starting materials consist of four green spheres and two purple spheres. The products consist of four green spheres and two purple spheres. This does not violate any of Dalton’s postulates: Atoms are neither created nor destroyed, but are redistributed in small, whole-number ratios.
#<hr>
problem = *2.2 Laws of Definite and Multiple Proportions
<h3>Example 2.2: Laws of Definite and Multiple Proportions</h3>
Laws of Definite and Multiple Proportions
#A sample of compound A (a clear, colorless gas) is analyzed and found to contain 4.27 g carbon and 5.69 g oxygen. A sample of compound B (also a clear, colorless gas) is analyzed and found to contain 5.19 g carbon and 13.84 g oxygen. Are these data an example of the law of definite proportions, the law of multiple proportions, or neither? What do these data tell you about substances A and B?
m[C]A = 4.27 g
m[O]A = 5.69 g
m[C]B = 5.19 g
m[O]B = 13.84 g
#<h4>Plan</h4>1) Calculate the mass ratios of compounds A and B. 2) See if the ratios are related
In compound A, the mass ratio of carbon to oxygen is:
r_A = m[C]A / m[O]A
In compound B, the mass ratio of carbon to oxygen is:
r_B = m[C]B / m[O]B
#The ratio of these ratios is:
r_AB = r_A / r_B
#This supports the law of multiple proportions. This means that A and B are different compounds, with A having one-half as much carbon per amount of oxygen (or twice as much oxygen per amount of carbon) as B. A possible pair of compounds that would fit this relationship would be A = CO2 and B = CO.
#<h4>Check Your Learning</h4>A sample of compound X (a clear, colorless, combustible liquid with a noticeable odor) is analyzed and found to contain 14.13 g carbon and 2.96 g hydrogen. A sample of compound Y (a clear, colorless, combustible liquid with a noticeable odor that is slightly different from X’s odor) is analyzed and found to contain 19.91 g carbon and 3.34 g hydrogen. Are these data an example of the law of definite proportions, the law of multiple proportions, or neither? What do these data tell you about substances X and Y?
#
#Answer:
#In compound X, the mass ratio of carbon to hydrogen is 14.13 g C: 2.96 g H. In compound Y, the mass ratio of carbon to oxygen is 19.91 g C: 3.34 g H. The ratio of these ratios is 14.13 g / 2.96 g / (19.91 g / 3.34 g) = 4.77 / 5.96 = 0.800 = 4/5. This small, whole-number ratio supports the law of multiple proportions. This means that X and Y are different compounds.
#<hr>
problem = 2.3 Composition of an Atom
<h3>Example 2.3: Composition of an Atom</h3>
Composition of an Atom
#Iodine is an essential trace element in our diet; it is needed to produce thyroid hormone. Insufficient iodine in the diet can lead to the development of a goiter, an enlargement of the thyroid gland (Figure 2.12).
#
#
#
#
#(a) Insufficient iodine in the diet can cause an enlargement of the thyroid gland called a goiter. (b) The addition of small amounts of iodine to salt, which prevents the formation of goiters, has helped eliminate this concern in the US where salt consumption is high. (credit a: modification of work by “Almazi”/Wikimedia Commons; credit b: modification of work by Mike Mozart)
#
#The addition of small amounts of iodine to table salt (iodized salt) has essentially eliminated this health concern in the United States, but as much as 40% of the world’s population is still at risk of iodine deficiency. The iodine atoms are added as anions, and each has a 1− charge and a mass number of 127. Determine the numbers of protons, neutrons, and electrons in one of these iodine anions.
#SolutionThe atomic number of iodine (53) tells us that a neutral iodine atom contains 53 protons in its nucleus and 53 electrons outside its nucleus. Because the sum of the numbers of protons and neutrons equals the mass number, 127, the number of neutrons is 74 (127 − 53 = 74). Since the iodine is added as a 1− anion, the number of electrons is 54 [53 – (1–) = 54].Check Your LearningAn ion of platinum has a mass number of 195 and contains 74 electrons. How many protons and neutrons does it contain, and what is its charge?
#Answer:
#78 protons; 117 neutrons; charge is 4+
#<hr>
problem = 2.4 Calculation of Average Atomic Mass
<h3>Example 2.4: Calculation of Average Atomic Mass</h3>
Calculation of Average Atomic Mass
#A meteorite found in central Indiana contains traces of the noble gas neon picked up from the solar wind during the meteorite’s trip through the solar system. Analysis of a sample of the gas showed that it consisted of 91.84% 20Ne (mass 19.9924 amu), 0.47% 21Ne (mass 20.9940 amu), and 7.69% 22Ne (mass 21.9914 amu). What is the average mass of the neon in the solar wind?
#<h4>Plan</h4>average mass=(0.9184×19.9924 amu)+(0.0047×20.9940 amu)+(0.0769×21.9914 amu)=(18.36+0.099+1.69)amu=20.15 amu
#
#The average mass of a neon atom in the solar wind is 20.15 amu. (The average mass of a terrestrial neon atom is 20.1796 amu. This result demonstrates that we may find slight differences in the natural abundance of isotopes, depending on their origin.)
#<h4>Check Your Learning</h4>A sample of magnesium is found to contain 78.70% of 24Mg atoms (mass 23.98 amu), 10.13% of 25Mg atoms (mass 24.99 amu), and 11.17% of 26Mg atoms (mass 25.98 amu). Calculate the average mass of a Mg atom.
#
#Answer:
#24.31 amu
#<hr>
problem = 2.5 Calculation of Percent Abundance
<h3>Example 2.5: Calculation of Percent Abundance</h3>
Calculation of Percent Abundance
#Naturally occurring chlorine consists of 35Cl (mass 34.96885 amu) and 37Cl (mass 36.96590 amu), with an average mass of 35.453 amu. What is the percent composition of Cl in terms of these two isotopes?
#<h4>Plan</h4>The average mass of chlorine is the fraction that is 35Cl times the mass of 35Cl plus the fraction that is 37Cl times the mass of 37Cl.