Answer:
Keep temperature constant and increase the pressure of the reaction. The rate of reaction increases.
Explanation:
First of all, the question is asking us to design an experiment to investigate the effect of pressure on the rate of reaction hence the pressure can not be held constant since it is the variable under investigation. This eliminates the first option.
Secondly, increasing the pressure of the reaction means that particles of the gas collide more frequently leading to a greater number of effective collisions and a consequent increase in the rate of reaction according to the collision theory.
Hence the answer above.
Answer:
53.6 g of N₂H₄
Explanation:
The begining is in the reaction:
N₂(g) + 2H₂(g) → N₂H₄(l)
We determine the moles of each reactant:
59.20 g / 28.01 g/mol = 2.11 moles of nitrogen
6.750 g / 2.016 g/mol = 3.35 moles of H₂
1 mol of N₂ react to 2 moles of H₂
Our 2.11 moles of N₂ may react to (2.11 . 2) /1 = 4.22 moles of H₂, but we only have 3.35 moles. The hydrogen is the limiting reactant.
2 moles of H₂ produce at 100 % yield, 1 mol of hydrazine
Then, 3.35 moles, may produce (3.35 . 1)/2 = 1.67 moles of N₂H₄
Let's convert the moles to mass:
1.67 mol . 32.05 g/mol = 53.6 g
Answer:- 1.62 moles
Solution:- At constant temperature and pressure, volume is directly proportional to the moles of the gas.
from given data, 
= 5.17 L, 
= 1.05 moles
= 8.00 L, 
= ?
Let's plug in the values in the formula:
On cross multiply:
= 1.62 moles
So, now the toy contains 1.62 moles of the air.
<span>The vaporization of br2 from liquid to gas state requires 7.4 k/cal /mol.</span>
