Answer:
Freezing point = 1.25
Explanation:
If we increase the concentration of the solution, the concentration of H+ does not change.
Convert 2.5% in to decimal
2.5% = 2.5 ÷100
= 0.025
The freezing point = 0.025 × 50
= 1.25
Answer:
1) Maximun ammount of nitrogen gas: 
2) Limiting reagent: 
3) Ammount of excess reagent: 
Explanation:
<u>The reaction </u>

Moles of nitrogen monoxide
Molecular weight: 


Moles of hydrogen
Molecular weight: 


Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess
1) <u>Maximun ammount of nitrogen gas</u> => when all NO reacted


2) <u>Limiting reagent</u>:
3) <u>Ammount of excess reagent</u>:


First, let's compute the number of moles in the system assuming ideal gas behavior.
PV = nRT
(663 mmHg)(1atm/760 mmHg)(60 L) = n(0.0821 L-atm/mol-K)(20+273 K)
Solving for n,
n = 2.176 moles
At standard conditions, the standard molar volume is 22.4 L/mol. Thus,
Standard volume = 22.4 L/mol * 2.176 mol =<em> 48.74 L</em>
It will result in an increase in the rate of rxn
Two types of stoichiometry are; molar mass and coefficients from balanced equation.