<span>CH4 + 4 Cl2 → CCl4 + 4 HCl
(4.00 mol CH4) x (1/1) x (0.70) = 2.80 mol CCl4
(4.00 mol CH4) x (4/1) x (0.70) = 11.2 mol HCl
CCl4 + 2 HF → CCl2F2 + 2 HCl
(2.80 mol CCl4) x (2/1) x (0.70) = 3.92 mol HCl
11.2 mol + 3.92 mol = 15.1 mol HCl from both steps</span>
The answer for this issue is:
The chemical equation is: HBz + H2O <- - > H3O+ + Bz-
Ka = 6.4X10^-5 = [H3O+][Bz-]/[HBz]
Let x = [H3O+] = [Bz-], and [HBz] = 0.5 - x.
Accept that x is little contrasted with 0.5 M. At that point,
Ka = 6.4X10^-5 = x^2/0.5
x = [H3O+] = 5.6X10^-3 M
pH = 2.25
(x is without a doubt little contrasted with 0.5, so the presumption above was OK to make)
We are given –
- Mass of
is 57.1 g and we are asked to find number of moles present in 57.1 g of
____________________
Now,Let's calculate the number of moles present in 57.1 g of
__________________________________
Answer: options B,D and F
Explanation:
Since redox reactions are those which involves both oxidation and reduction
In B , Cu is oxidized and S gets reduced
D, Na gets oxidized and hydrogen gets reduced
F, carbon gets oxidized and Oxygen gets reduced
In g, there is no change in oxidation no of s in both product and Reactants is same +4
Similarly in the case of Ag and Mg.
