Answer:
Option A (0.043 g) is the correct answer.
Explanation:
Given:
= 43 mg
As we know,

then,
⇒ 
Thus, the above is the correct alternative.
A redox reaction --> a reaction whereby oxidation & reduction occurs
Reduction:
Charge of Cl2 = 0
Charge of Cl- in NaCl = -1
Hence, since charge of Cl2 decreased from 0 in Cl2 to -1 in NaCl, reduction occured.
Oxidation:
Charge of Na = 0
Charge of Na+ in NaCl = +1
Hence, since charge of Na increased from 0 in Na to +1 in NaCl, oxidation occured.
Since both oxidation & reduction occured in the reaction, it is a redox reaction.
<u>Answer:</u> The equilibrium concentration of
is 0.332 M
<u>Explanation:</u>
We are given:
Initial concentration of
= 2.00 M
The given chemical equation follows:

<u>Initial:</u> 2.00
<u>At eqllm:</u> 2.00-2x x x
The expression of
for above equation follows:
![K_c=\frac{[CO_2][CF_4]}{[COF_2]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO_2%5D%5BCF_4%5D%7D%7B%5BCOF_2%5D%5E2%7D)
We are given:

Putting values in above expression, we get:

Neglecting the value of x = 1.25 because equilibrium concentration of the reactant will becomes negative, which is not possible
So, equilibrium concentration of ![COF_2=(2.00-2x)=[2.00-(2\times 0.834)]=0.332M](https://tex.z-dn.net/?f=COF_2%3D%282.00-2x%29%3D%5B2.00-%282%5Ctimes%200.834%29%5D%3D0.332M)
Hence, the equilibrium concentration of
is 0.332 M
Density (d) which is the quotient when mass (m) is divided by volume (v) is usually reported in terms of g/mL.
d = m /v
Substituting the known values,
d = (1.62 kg) x (1000 g/ 1 kg) / (205 mL)
The answer would be approximately equal to 7.9 g/mL.