Answer:
The number of moles of CaCO3 on the bag is 112.90 moles
Explanation:
number mole (n) = mass (m) divided by molecular mass (Mm)
Mm of CaCO3 = 100.0869 g/mole
mass in grams = 11.3 Kg x (10^3 g/1 Kg) = 11300 grams
number of moles (n) = 11300 grams divided by 100.0869 grams per mole = 112.90 moles of CaCO3 in the bag.
Answer:
b- 4.4 * 10^-12.
Explanation:
Hello.
In this case, as the reaction:
A + 2B → 3C
Has an equilibrium expression of:
![K_1=\frac{[C]^3}{[A][B]^2}=2.1x10^{-6}](https://tex.z-dn.net/?f=K_1%3D%5Cfrac%7B%5BC%5D%5E3%7D%7B%5BA%5D%5BB%5D%5E2%7D%3D2.1x10%5E%7B-6%7D)
If we analyze the reaction:
2A + 4B → 6C
Which is twice the initial one, the equilibrium expression is:
![K_2=\frac{[C]^6}{[A]^2[B]^4}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BC%5D%5E6%7D%7B%5BA%5D%5E2%5BB%5D%5E4%7D)
It means that the equilibrium constant of the second reaction is equal to the equilibrium constant of the first reaction powered to second power:

Thus, the equilibrium constant of the second reaction turns out:

Therefore, the answer is b- 4.4 * 10^-12.
Best regards.
Answer:-
,
, ![[CO_3^2^-]=0.254M](https://tex.z-dn.net/?f=%5BCO_3%5E2%5E-%5D%3D0.254M)
Solution:- We are asked to calculate the molarity of sodium carbonate solution as well as the sodium and carbonate ions.
Molarity is moles of solute per liter of solution. We have been given with 6.73 grams of sodium carbonate and the volume of solution is 250.mL. Grams are converted to moles and mL are converted to L and finally the moles are divided by liters to get the molarity of sodium carbonate.
Molar mass of sodium carbonate is 105.99 gram per mol. The calculations for the molarity of sodium carbonate are shown below:

= 
So, molarity of sodium carbonate solution is 0.254 M.
sodium carbonate dissociate to give the ions as:

There is 1:2 mol ratio between sodium carbonate and sodium ion. So, the molarity of sodium ion will be two times of sodium carbonate molarity.
= 0.508 M
There is 1:1 mol ratio between sodium carbonate and carbonate ion. So, the molarity of carbonate ion will be equal to the molarity of sodium carbonate.
![[CO_3^2^-]=0.254M](https://tex.z-dn.net/?f=%5BCO_3%5E2%5E-%5D%3D0.254M)
Answer is: 0.102 moles of HCl would react.
Balanced chemical reaction:
2HCl(aq) + Sr(OH)₂ → SrCl₂(aq) + 2H₂O(l).
V(Sr(OH)₂) = 37.1 mL ÷ 1000 mL/L.
V(Sr(OH)₂) = 0.0371 L; volume of the strontium hydroxide solution.
c(Sr(OH)₂) = 0.138 M; molarity of the strontium hydroxide solution.
n(Sr(OH)₂) = c(Sr(OH)₂) · V(Sr(OH)₂).
n(Sr(OH)₂) = 0.0371 L · 0.138 mol/L.
n(Sr(OH)₂) = 0.0051 mol; amount of the strontium hydroxide.
From balanced chemical reaction: n(Sr(OH)₂) : n(HCl) = 1 : 2.
n(HCl) = 2 · n(Sr(OH)₂).
n(HCl) = 2 · 0.0051 mol.
n(HCl) = 0.0102 mol; amount of the hydrochloric acid.