Answer:47.05% is the percent yield of nitrogen in the reaction.
Explanation:
heoretical yield of nitrogen gas = x
Moles of ammonia =
According to reaction,2 moles of ammonia gives 1 mol of nitrogen gas.
Then 2.3529 mol of ammonia will give:
of nitrogen gas
Mass of 1.1764 moles of nitrogen gas,x = 1.1764 mol × 28 g/mol=32.94 g
Experiential yield of nitrogen gas = 15.5 g
Percentage yield:
hope that help
47.05% is the percent yield of nitrogen in the reaction.
44.0095 you're welcome hope this helps
Answer:
https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=2ahUKEwjkwv-cqrjnAhVCheAKHWaFBBgQFjAAegQICBAB&url=https%3A%2F%2Fwww.acs.org%2Fcontent%2Fdam%2Facsorg%2Feducation%2Fresources%2Fk-8%2Finquiryinaction%2Fstudent-activity-sheets%2Fgrade-5%2Fchapter-3%2Flesson-3.3-forming-a-precipitate.pdf&usg=AOvVaw1fT7fpXG9PNWroM87puvgQ
Explanation:
that has the answers copy and paste it in your google
Answer:
Explanation:
Hello,
In this case, for a concentration of 0.42 M of benzoic acid whose Ka is 6.3x10⁻⁵ in 0.33 M sodium benzoate, we use the Henderson-Hasselbach equation to compute the required pH:
![pH=pKa+log(\frac{[base]}{[acid]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29)
Whereas the concentration of the base is 0.33 M and the concentration of the acid is 0.42 M, thereby, we obtain:
![pH=-log(Ka)+log(\frac{[base]}{[acid]} )\\\\pH=-log(6.3x10^{-5})+log(\frac{0.33M}{0.42M} )\\\\pH=4.1](https://tex.z-dn.net/?f=pH%3D-log%28Ka%29%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29%5C%5C%5C%5CpH%3D-log%286.3x10%5E%7B-5%7D%29%2Blog%28%5Cfrac%7B0.33M%7D%7B0.42M%7D%20%29%5C%5C%5C%5CpH%3D4.1)
Regards.
