Answer:

Explanation:
= Initial pressure = 931 torr = 
= Final pressure = 113 kPa
= Initial volume = 350 mL
= Final volume
From the Boyle's law we have

The volume the gas would occupy is
.
According to Balance chemical equation,
N₂ + 3 H₂ → 2 NH₃
1 mole of Nitrogen reacts with 3 moles of Hydrogen to produce 2 mole of Ammonia.
It is known that i mole of any gas at standard temperature and pressure occupies 22.4 L of Volume. So, we can also say,
22.4 L (1 × 22.4) of Nitrogen gas (in question it is taken in excess) reacts with 67.2 L (22.4 × 3) of Hydrogen gas to produce 44.8 L (22.4 × 2) of Ammonia.
Result:
44.8 L is the correct answer.
When a substance is entering a phase change, the gain or loss of heat is a result of energy gained or lost in forming or breaking intermolecular interaction.
The constant temperatures occur when a substance is undergoing a phase transition. If heat is removed from a substance , such as in freezing and condensation , then the process is exothermic . In this instance , heat is decreasing the speed of the molecules causing then move slower.
Example : liquid to solid and gas to liquid .
These changes release heat to the surrounding.
To learn more about phase change,
brainly.com/question/12390797
#SPJ4
The answer would be balanced
Answer:
ΔG° of reaction = -47.3 x
J/mol
Explanation:
As we can see, we have been a particular reaction and Energy values as well.
ΔG° of reaction = -30.5 kJ/mol
Temperature = 37°C.
And we have to calculat the ΔG° of reaction in the biological cell which contains ATP, ADP and HPO4-2:
The first step is to calculate the equilibrium constant for the reaction:
Equilibrium Constant K = ![\frac{[HPO4-2] x [ADP]}{ATP}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BHPO4-2%5D%20x%20%5BADP%5D%7D%7BATP%7D)
And we have values given for these quantities in the biological cell:
[HP04-2] = 2.1 x
M
[ATP] = 1.2 x
M
[ADP] = 8.4 x
M
Let's plug in these values in the above equation for equilibrium constant:
K = ![\frac{[2.1x10^{-3}] x [8.4x10^{-3}] }{[1.2 x 10^{-2}] }](https://tex.z-dn.net/?f=%5Cfrac%7B%5B2.1x10%5E%7B-3%7D%5D%20x%20%5B8.4x10%5E%7B-3%7D%5D%20%7D%7B%5B1.2%20x%2010%5E%7B-2%7D%5D%20%7D)
K = 1.47 x
M
Now, we have to calculate the ΔG° of reaction for the biological cell:
But first we have to convert the temperature in Kelvin scale.
Temp = 37°C
Temp = 37 + 273
Temp = 310 K
ΔG° of reaction = (-30.5
) + (8.314)x (310K)xln(0.00147)
Where 8.314 = value of Gas Constant
ΔG° of reaction = (-30.5 x
) + (-16810.68)
ΔG° of reaction = -47.3 x
J/mol