Answer:
The fraction of energy used to increase the internal energy of the gas is 0.715
Explanation:
Step 1: Data given
Cv for nitrogen gas = 20.8 J/K*mol
Cp for nitrogen gas = 29.1 J/K*mol
Step 2:
At a constant volume, all the heat will increase the internal energy of the gas.
At constant pressure, the gas expands and does work., if the volume changes.
Cp= Cv + R
⇒The value needed to change the internal energy is shown by Cv
⇒The work is given by Cp
To find what fraction of the energy is used to increase the internal energy of the gas, we have to calculate the value of Cv/Cp
Cv/Cp = 20.8 J/K*mol / 29.1 J/K*mol
Cv/Cp = 0.715
The fraction of energy used to increase the internal energy of the gas is 0.715
The heat of combustion for methanol is 727 kj/mol
<em><u>calculation</u></em>
calculate the moles of methanol (CH3OH)
moles = mass/molar mass
molar mass of methanol = 12 +( 1 x3) +16 + 1= 32 g /mol
moles is therefore= 64.0 g / 32 g/mol = 2 moles
Heat of combustion is therefore = 1454 Kj / 2 moles = 727 Kj/mol
Answer:
There are
17.01
Explanation:
The chemical formula for calcium phosphate is
Ca
3
(PO
4
)
2
. This means that in one mole of calcium phosphate, there are three calcium ions and two phosphate ions.
The equilibrium constant of a reaction is defined as:
"The ratio between equilibrium concentrations of products powered to their reaction quotient and equilibrium concentration of reactants powered to thier reaction quotient".
The reaction quotient, Q, has the same algebraic expressions but use the actual concentrations of reactants.
To solve this question we need this additional information:
<em>For this reaction, K = 6.0x10⁻² and the initial concentrations of the reactants are:</em>
<em>[N₂] = 4.0M; [NH₃] = 1.0x10⁻⁴M and [H₂] = 1.0x10⁻²M</em>
<em />
Thus, for the reaction:
N₂ + 3H₂ ⇄ 2NH₃
The equilibrium constant, K, of this reaction, is defined as:
![K = 6.0x10^{-2} = \frac{[NH_3]^2}{[N_2][H_2]^3}](https://tex.z-dn.net/?f=K%20%3D%206.0x10%5E%7B-2%7D%20%3D%20%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5BH_2%5D%5E3%7D)
And Q, is:
![Q = \frac{[NH_3]^2}{[N_2][H_2]^3}](https://tex.z-dn.net/?f=Q%20%3D%20%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5BH_2%5D%5E3%7D)
Where actual concentrations are:
[NH₃] = 1.0x10⁻⁴M
[N₂] = 4.0M
[H₂] = 2.5x10⁻¹M
Replacing:
![Q = \frac{[1.0x10^{-4}]^2}{[4.0][2.5x10^{-1}]^3}](https://tex.z-dn.net/?f=Q%20%3D%20%5Cfrac%7B%5B1.0x10%5E%7B-4%7D%5D%5E2%7D%7B%5B4.0%5D%5B2.5x10%5E%7B-1%7D%5D%5E3%7D)
<h3>Q = 1.6x10⁻⁷</h3>
As Q < K,
<h3>The chemical system will shift to the right in order to produce more NH₃</h3>
Learn more about chemical equililbrium in:
brainly.com/question/24301138