Answer: The value of the equilibrium constant, Kc, for the reaction is 0.013
Explanation:
Initial concentration of
= 3.60 M
Initial concentration of
= 3.60 M
The given balanced equilibrium reaction is,

Initial conc. 3.60 M 3.60 M 0 M 0 M
At eqm. conc. (3.60-4x) M (3.60-7x) M (2x) M (6x) M
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[N_2O_4]^2\times [H_2O]^6}{[NH_3]^4\times[Cl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BN_2O_4%5D%5E2%5Ctimes%20%5BH_2O%5D%5E6%7D%7B%5BNH_3%5D%5E4%5Ctimes%5BCl_2%5D%7D)
![[N_2O_4]=0.60M](https://tex.z-dn.net/?f=%5BN_2O_4%5D%3D0.60M)
2x = 0.60 M
x= 0.30 M
Now put all the given values in this expression, we get :


Answer:
- 602 mg of CO₂ and 94.8 mg of H₂O
Explanation:
The<em> yield</em> is measured by the amount of each product produced by the reaction.
The chemical formula of <em>fluorene</em> is C₁₃H₁₀, and its molar mass is 166.223 g/mol.
The <em>oxidation</em>, also know as combustion, of this hydrocarbon is represented by the following balanced chemical equation:

To calculate the yield follow these steps:
<u>1. Mole ratio</u>

<u />
<u>2. Convert 175mg of fluorene to number of moles</u>
- Number of moles = mass in grams / molar mass
<u>3. Set a proportion for each product of the reaction</u>
a) <u>For CO₂</u>
i) number of moles


ii) mass in grams
The molar mass of CO₂ is 44.01g/mol
- mass = number of moles × molar mass
- mass = 0.013686 moles × 44.01 g/mol = 0.602 g = 602mg
b) <u>For H₂O</u>
i) number of moles

ii) mass in grams
The molar mass of H₂O is 18.015g/mol
- mass = number of moles × molar mass
- mass = 0.00526 moles × 18.015 g/mol = 0.0948mg = 94.8 mg
<u>Answer:</u>
16.8 cups
<u>Explanation:</u>
We are given that there are 4.2 cups in in 1 liter of a liquid and we are to find out the number of cups that there will be in 4 liters.
To find this out, we will use the ratio method.
We know that:
1 L = 4.2 cups
So we can write it as:



Therefore, there are 16.8 cups in 4 liters.
Explanation:
lactic acid because of the acid it will release in the cellular respiration