Problem Set 3 Answer Key (PDF)




File information


Author: Robert Bauer

This PDF 1.5 document has been generated by Microsoftร‚ยฎ Word 2013, and has been sent on pdf-archive.com on 10/09/2016 at 03:09, from IP address 50.164.x.x. The current document download page has been viewed 996 times.
File size: 278.44 KB (1 page).
Privacy: public file




Document preview - Problem Set 3 Answer Key.pdf - Page 1/1





File preview


Chemistry Lesson: Steemit Edition Problem Set 3

1.) How many neutrons are present in 54
24๐ถ๐‘Ÿ ? (refer to Chemistry Lesson: Part 3 if you do not
remember how to determine that)

There are 30 neutrons in that Chromium atom. Remember molecular mass is that of protons + neutrons,
and the atomic number is the number of protons. To calculate number of neutrons we subtract the
atomic number from the atomic mass (54 - 24 = 30).

2.) What is the empirical formula for a compound with the following composition: 54.5 % C, 13.7 %
H, and 31.8 % N? (Bonus: the molecular mass of this compound is ~88 g/mol, the first person to
post the correct molecular formula and name of the compound in the comments of Chemistry
Lesson: Part 4 will get 5 SBD as a reward).
1 ๐‘š๐‘œ๐‘™
54.5 % ๐ถ โˆ— 100๐‘” = 54.5 ๐‘” ๐ถ โˆ—
= 4.54 ๐‘€๐‘œ๐‘™ ๐ถ
12.01 ๐‘” ๐ถ
1 ๐‘š๐‘œ๐‘™
13.7 % ๐ป โˆ— 100๐‘” = 13.7 ๐‘” ๐ป โˆ—
= 13.59 ๐‘š๐‘œ๐‘™ ๐ป
1.008 ๐‘” ๐ถ
1 ๐‘š๐‘œ๐‘™
31.8 % ๐‘ โˆ— 100๐‘” = 31.8 ๐‘” ๐‘ โˆ—
= 2.26 ๐‘š๐‘œ๐‘™ ๐‘
14.01 ๐‘” ๐ถ
Nitrogen is the smallest # of moles so we divide each number by 2.26. 4.54/2.26 = 2C.
13.59/2.26 = 6 H, 2.26/2.26 = 1 N. So our empirical formula is C2H6N.

If you want the bonusโ€ฆ youโ€™ve got to figure that one out on your own.
3.) Balance the following chemical reaction: Al + H2SO4 ๏ƒ  Al2(SO4)3 + H2
This is the reaction of aluminum with concentrated sulfuric acid, if you are curious what this
reaction looks like see the following YouTube video (itโ€™s NOT a crazy one):
(https://www.youtube.com/watch?v=vR2ivraWTyU)
2Al + H2SO4 ๏ƒ  Al2(SO4)3 + H2
2Al + 3H2SO4 ๏ƒ  Al2(SO4)3 + H2
3Al + 3H2SO4 ๏ƒ  Al2(SO4)3 + 3H2
Balanced Aluminum because it was the easiest first. Then balanced SO4, there were 3 on the
product side (Compound)number of units so added 3 to the H2SO4 to make that equal. Finally
balanced the hydrogens. We had 6 on the reactants side, so added the coefficient of 3 to the H2
to put 6 on the product side.






Download Problem Set 3 Answer Key



Problem Set 3 Answer Key.pdf (PDF, 278.44 KB)


Download PDF







Share this file on social networks



     





Link to this page



Permanent link

Use the permanent link to the download page to share your document on Facebook, Twitter, LinkedIn, or directly with a contact by e-Mail, Messenger, Whatsapp, Line..




Short link

Use the short link to share your document on Twitter or by text message (SMS)




HTML Code

Copy the following HTML code to share your document on a Website or Blog




QR Code to this page


QR Code link to PDF file Problem Set 3 Answer Key.pdf






This file has been shared publicly by a user of PDF Archive.
Document ID: 0000481952.
Report illicit content