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:
0.093 mole of C₆H₁₂.
Explanation:
We'll begin by calculating the molar mass of C₆H₁₂. This can be obtained as follow:
Molar mass of C₆H₁₂ = (12×6) + (12×1)
= 72 + 12
= 84 g/mol
Finally, we shall determine the number of mole in 7.8 g of C₆H₁₂. This can be obtained as follow:
Molar mass of C₆H₁₂ = 84 g/mol
Mass of C₆H₁₂ = 7.8 g
Mole of C₆H₁₂ =?
Mole = mass / molar mass
Mole of C₆H₁₂ = 7.8 / 84
Mole of C₆H₁₂ = 0.093 mole
Thus, 7.8 g contains 0.093 mole of C₆H₁₂.
Answer:
The reaction quotient (Q) before the reaction is 0.32
Explanation:
Being the reaction:
aA + bB ⇔ cC + dD
![Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BC%5D%5E%7Bc%7D%20%2A%5BD%5D%5E%7Bd%7D%20%7D%7B%5BA%5D%5E%7Ba%7D%2A%5BB%5D%5E%7Bb%7D%20%20%7D)
where Q is the so-called reaction quotient and the concentrations expressed in it are not those of the equilibrium but those of the different reagents and products at a certain instant of the reaction.
The concentration will be calculated by:

You know the reaction:
PCl₅ (g) ⇌ PCl₃(g) + Cl₂(g).
So:
![Q=\frac{[PCl_{3} ] *[Cl_{2} ] }{[PCl_{5} ]}](https://tex.z-dn.net/?f=Q%3D%5Cfrac%7B%5BPCl_%7B3%7D%20%5D%20%2A%5BCl_%7B2%7D%20%5D%20%7D%7B%5BPCl_%7B5%7D%20%5D%7D)
The concentrations are:
- [PCl₃]=

- [Cl₂]=

- [PCl₅]=

Replacing:

Solving:
Q= 0.32
<u><em>The reaction quotient (Q) before the reaction is 0.32</em></u>
Answer:
14 gallons
Explanation:
357 divided by 14 = 25.5 and if you check your answer 14 x 25.5 = 357 (i hope this is right)