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# AP CHEM HOMEWORK HELP PLZ BEST ANSWER GETS 10 POINTS?

Topic: Ap chem homework
July 19, 2019 / By Gennie
Question: a 0.573 mg sample ibuprofen, which is the active ingredient in Advil, gives 1.59 mg of CO2 and 0.4504 mg of H2O on combustion. if ibuprofen contains only C,H, and O what is the percent composition of ibuprofen. help on how to do and what equations will be very helpful.

## Best Answers: AP CHEM HOMEWORK HELP PLZ BEST ANSWER GETS 10 POINTS?

Delight | 4 days ago
This exact question with the asnwer is on www.myAPtrade.com. That website has everysingle released AP exam with solutions, test banks for AP classes and lab reports. It costs like 25\$ but its worth it..exact questions to my midterm for AP Chem were on there. Step 1: Find mg of C (1.59 mg CO2) / (44.0096 g CO2/mol) x (1/1) x (12.0108 g C/mol) = 0.43393 mg C Step 2: Find mg of H (0.4504 mg H2O) / (18.0153 g H2O/mol) x (2/1) x (1.0079 g H/mol) = 0.050397 mg H Step 3: Find mg of O (0.573 mg) - (0.43393 mg C) - (0.050397 mg H) = 0.088673 mg O Step 4: Divide mg of specific element by total mg to get % composition of each element (0.43393 mg C) / (0.573 mg) = 0.757 = 75.7% C (0.050397 mg H) / (0.573 mg) = 0.0880 = 8.80% H (0.088673 mg O) / (0.573 mg) = 0.155 = 15.5% O
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Ok, bribery helps. Each gamma ray will have an energy of 511 KeV. (511000 eV) To convert this to kJ/Mol, you need to know that 1 eV = 96.485 kJ/Mol. So, each gamma ray will have 511000 X 96.485 kJ/Mol = 4.93E+007 kJ/Mol Since two gamma rays are generated, just double it to 9.86E+007 kJ/Mol Check the Wiki for Electron Volt to see a more detailed discussion.
using fact the floor gets less warm and freezes the water molecules boost while the water expands, the hoses will burst or boost small holes in it. If it would not have a freeze spigot, it additionally will freeze as much as the warmth wall or section, and burst.

Bryanne
(1.59 mg CO2) / (44.0096 g CO2/mol) x (1/1) x (12.0108 g C/mol) = 0.43393 mg C (0.4504 mg H2O) / (18.0153 g H2O/mol) x (2/1) x (1.0079 g H/mol) = 0.050397 mg H (0.573 mg) - (0.43393 mg C) - (0.050397 mg H) = 0.088673 mg O (0.43393 mg C) / (0.573 mg) = 0.757 = 75.7% C (0.050397 mg H) / (0.573 mg) = 0.0880 = 8.80% H (0.088673 mg O) / (0.573 mg) = 0.155 = 15.5% O
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Alleen
(a million.fifty nine mg CO2) / (forty four.0096 g CO2/mol) x (a million/a million) x (12.0108 g C/mol) = zero.43393 mg C (zero.4504 mg H2O) / (18.0153 g H2O/mol) x (two/a million) x (a million.0079 g H/mol) = zero.050397 mg H (zero.573 mg) - (zero.43393 mg C) - (zero.050397 mg H) = zero.088673 mg O (zero.43393 mg C) / (zero.573 mg) = zero.757 = seventy five.7% C (zero.050397 mg H) / (zero.573 mg) = zero.0880 = eight.eighty% H (zero.088673 mg O) / (zero.573 mg) = zero.a hundred and fifty five = 15.five% O
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