Answer:
2.2mL
Explanation:
First, let us analyse what was given from the question:
C1 = 2.09M
V1 =?
C2 = 0.046M
V2 =100mL
Using dilution formula (C1V1 = C2V2), we can calculate the volume of the original solution as follows:
C1V1 = C2V2
2.09 x V1 = 0.046 x 100
Divide both side by the coefficient of V1 ie 2.09, we have:
V1 = (0.046 x 100) / 2.09
V1 = 2.2mL
I believe it's D! Hope you can help with my question!
<u>Answer:</u> The equilibrium concentration of water is 0.597 M
<u>Explanation:</u>
Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as 
For a general chemical reaction:

The expression for
is written as:
![K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%7B%5BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
The concentration of pure solids and pure liquids are taken as 1 in the expression.
For the given chemical reaction:

The expression of
for above equation is:
![K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_2O%5D%5E2%7D%7B%5BH_2S%5D%5E2%5Ctimes%20%5BO_2%5D%7D)
We are given:
![[H_2S]_{eq}=0.671M](https://tex.z-dn.net/?f=%5BH_2S%5D_%7Beq%7D%3D0.671M)
![[O_2]_{eq}=0.587M](https://tex.z-dn.net/?f=%5BO_2%5D_%7Beq%7D%3D0.587M)

Putting values in above expression, we get:
![1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}](https://tex.z-dn.net/?f=1.35%3D%5Cfrac%7B%5BH_2O%5D%5E2%7D%7B%280.671%29%5E2%5Ctimes%200.587%7D)
![[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M](https://tex.z-dn.net/?f=%5BH_2O%5D%3D%5Csqrt%7B%281.35%5Ctimes%200.671%5Ctimes%200.671%5Ctimes%200.587%29%7D%3D0.597M)
Hence, the equilibrium concentration of water is 0.597 M
The reaction between Na2S and CuSO4 will give us the balanced chemical reaction of,
Na2S + CUSO4 --> Na2SO4 + CuS
This means that for every 78g of Na2S, there needs to be 159.6 g of CuSO4. The ratio is equal to 0.4887 of Na2S: 1 of CuSO4. Thus, for every 12.1g of CuSO4, we need only 5.91 g of Na2S. Thus, there is an excess of 9.58 g of Na2S. The answer is letter C.
Generally, chemists prefer to use morality (B) because it only invovles measuring the final volume of the solution and amount of moles of the solute
Hope this helps