Answer:
The Kc of this reaction is 311.97
Explanation:
Step 1: Data given
Kp = 0.174
Temperature = 243 °C
Step 2: The balanced equation
N2(g) + 3H2(g) ⇌ 2NH3(g)
Step 3: Calculate Kc
Kp = Kc *(RT)^Δn
⇒ with Kp = 0.174
⇒ with Kc = TO BE DETERMINED
⇒ with R = the gas constant = 0.08206 Latm/Kmol
⇒ with T = the temperature = 243 °C = 516 K
⇒ with Δn = number of moles products - moles reactants 2 – (1 + 3) = -2
0.174 = Kc (0.08206*516)^-2
Kc = 311.97
The Kc of this reaction is 311.97
Answer:
2KMnO4(aq) + 16HCl(aq) ------> 2MnCl2(aq) + 2KCl(aq) + 8H2O(l) + 5Cl2(g)
Explanation:
Chlorine is a diatomic halogen gas known for its greenish-yellow colour. It has a pungent smell and is only moderately soluble in water.
It is a very reactive gas and is never found in free state in nature.
Chlorine can be prepared in the laboratory by oxidation of hydrochloric acid using KMnO4 as follows;
2KMnO4(aq) + 16HCl(aq) ------> 2MnCl2(aq) + 2KCl(aq) + 8H2O(l) + 5Cl2(g)
The set up does not need to be heated.
2.4212 X 10^ 7
How I at least figure this problem out is I take a pencil and start on the right side of the 0 and make a loop to the left for each number and count until I get to the first two numbers that are between 1-9 when reading from left to right. This is where you put the decimal point. Some teachers rather you keep the 0's there, while others prefer one to get rid of them. Anyways with that new decimal number, you multiply the decimal by ten to what ever number you counted, which was 7.
When we have:
Zn(OH)2 → Zn2+ 2OH- with Ksp = 3 x 10 ^-16
and:
Zn2+ + 4OH- → Zn(OH)4 2- with Kf = 2 x 10^15
by mixing those equations together:
Zn(OH)2 + 2OH- → Zn(OH)4 2- with K = Kf *Ksp = 3 x 10^-16 * 2x10^15 =0.6
by using ICE table:
Zn(OH)2 + 2OH- → Zn(OH)4 2-
initial 2m 0
change -2X +X
Equ 2-2X X
when we assume that the solubility is X
and when K = [Zn(OH)4 2-] / [OH-]^2
0.6 = X / (2-2X)^2 by solving this equation for X
∴ X = 0.53 m
∴ the solubility of Zn(OH)2 = 0.53 M
