Answer:
C₂H₂ + 3H₂ ⟶ 2CH₄
Explanation:
The initial concentrations are:
[CH₄] = 6.30 ÷ 6.00 = 1.05 mol·L⁻¹
[C₂H₂] = 4.20 ÷ 6.00 = 0.700 mol·L⁻¹
[H₂] = 11.15 ÷ 6.00 = 1.858 mol·L⁻¹
2CH₄ ⇌ C₂H₂ + 3H₂
I/mol·L⁻¹: 1.05 0.700 1.858
![Q = \dfrac{\text{[C$_{2}$H$_{2}$][H$_{2}$]}^{3}}{\text{[CH$_{4}$]}^{2}} = \dfrac{ 0.700\times 1.858^{3}}{1.05^{2}}= 4.07](https://tex.z-dn.net/?f=Q%20%3D%20%5Cdfrac%7B%5Ctext%7B%5BC%24_%7B2%7D%24H%24_%7B2%7D%24%5D%5BH%24_%7B2%7D%24%5D%7D%5E%7B3%7D%7D%7B%5Ctext%7B%5BCH%24_%7B4%7D%24%5D%7D%5E%7B2%7D%7D%20%3D%20%5Cdfrac%7B%200.700%5Ctimes%201.858%5E%7B3%7D%7D%7B1.05%5E%7B2%7D%7D%3D%204.07)
Q > K
That means we have too many products.
The reaction will go to the left to get rid of the excess products.
C₂H₂ + 3H₂ ⟶ 2CH₄
The additional information required to calculate the cost of the drive has been distance between the two cities. Thus, option B is correct.
The distance traveled by the object set the cost per unit has been given by the product of the distance and the cost.
<h3>Detail required for calculation cost</h3>
The given question has known cost of the fuel per gallon.
The mileage of the car has been known.
The distance traveled by the car has not been known. Thus, to calculate the cost of the drive, the distance between the two points has to be known. Thus, option B is correct.
Learn more about distance traveled, here:
brainly.com/question/1214027
Answer:
Kc → 5.58×10⁻⁴
Explanation:
Equilibrium reaction is:
2NOCl (g) ⇄ 2NO (g) + Cl₂(g)
Initially we have 1.25 moles of NOCl
After the equilibrium, we have 1.10 moles. So, during the process:
(1.25 mol - 1.1 mol) = 0.15 moles have reacted.
As ratio are 2:2, and 2:1, 0.15 moles of NO and (0.15 /2) = 0.075 moles of chlorine, were produced in the equilibrium.
Finally in equilibrium we have: 1.10 moles of NOCl, 0.15 moles of NO and 0.075 moles of Cl₂. But these amount are not molar, so we need molar concentration in order to determine Kc:
1.10 mol /2.50L = 0.44 M
0.15 mol / /2.50L = 0.06 M
0.075 mol /2.50L = 0.03 M
Let's make expression for Kc → [Cl₂] . [NO]² / [NOCl]²
Kc = (0.03 . 0.06²) / 0.44² → 5.58×10⁻⁴
Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
![\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bn%5Ctimes%20H_f_%7Bproducts%7D%5D-%5Bn%5Ctimes%20H_f_%7Breactants%7D%5D)
Putting the values we get :
![\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B8%5Ctimes%20H_f_%7BCO_2%7D%2B10%5Ctimes%20H_f_%7BH_2O%7D%5D-%5B2%5Ctimes%20H_f_%7BC_4H_%7B10%7D%2B13%5Ctimes%20H_f_%7BO_2%7D%7D%5D)
![\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%288%5Ctimes%20-393.5%29%2B%2810%5Ctimes%20-241.82%29%5D-%5B%282%5Ctimes%20-125.7%29%2B%2813%5Ctimes%200%29%5D)
2 moles of butane releases heat = 5314.8 kJ
1 mole of butane release heat = 
Thus enthalpy of combustion per mole of butane is -2657.4 kJ
Explanation:
Using the Combined Gas Law, which is:
<em>(With </em>
<em>being initial pressure, volume and temperature; and</em>
<em />
<em> being the new values)</em>
<em />
We can move the units around in order to solve for 
, which would look like this:
Then we convert the Celsius temperature to Kelvin:
°
= 
° = 
And now, we plug in all of the values and solve, with volume remaining as a constant:
<em>or </em>
