N2 + 3H2 ----> 2NH3
<span>you can see 3 moles H2 reacts to form 2 moles NH3 </span>
<span>Therefore moles NH3 = 2 / 3 x moles H2 </span>
<span>= 2/3 x 12.0 mol </span>
<span>= 8.00 mol NH3 hope this help</span>
Answer:
72.3 c
Explanation:the celcuis would over power the fairnheight
To calculate the molarity you only need to know the number of moles in the solution and the volume of that solution. This exercise gives both and with that you divide moles by volume(usually in liters).
500 ml equals 0,5 L
molarity= number of moles/ volume
molarity=0,75 x 0,5
= 0,375 mol/L
The bottom of a cricket wing is covered with teeth-like ridges that make<span> it rough. The upper surface of the wing is like a scraper. When </span>crickets<span> rub the upper and lower parts of their wings together, they create a chirping</span>sound<span> called “stridulating."</span>
Answer:
The percent by mass of copper in the mixture was 32%
Explanation:
The ammount of HNO₃ used is:
mol HNO₃ = volume * concentration
mol HNO₃ = 0.015 l * 15.8 mol/l
mol HNO₃ = 0.237 mol
According to the reaction, 4 mol HNO₃ will react with 1 mol Cu and produce 1 mol Cu²⁺. Since we have 0.237 mol HNO₃, the amount of Cu that could react would be (0.237 mol HNO₃ * 1 mol Cu / 4 mol HNO₃) 0.06 mol. This reaction would produce 0.060 mol Cu²⁺, however, only 0.010 mol Cu²⁺ were obtained, indicating that only 0.010 mol Cu were present in the mixture. This means that the acid was in excess, so we can assume that all copper present in the mixture has reacted.
Since 0.010 mol of Cu²⁺ were produced, the amount of Cu was 0.01 mol.
1 mol of Cu has a mass of 63.5 g, then 0.01 mol has a mass of:
0.01 mol Cu * 63.5 g / 1 mol = 0.635 g.
Since this amount was present in 2.00 g mixture, the amount of copper in 100 g of the mixture will be:
100 g(mixture) * 0.635 g Cu / 2 g(mixture) = 32 g
Then, the percent by mass of Cu (which is the mass of Cu in 100 g mixture) is 32%
