The mass of Zr deposited in the process is 41.4 g.
<h3>What is electrolytic cell?</h3>
An electrolytic cell is a chemical cell which produces electrical energy by non-spontaneous chemical processes.
From the question;
Zr^4+(aq) + 4e ------> Zr(s)
We know that;
91 g of Zr is deposited by 4(96500) C
xg of Zr is deposited by (7.92 × 6.16 × 60 × 60) C
xg = 91 g × (7.92 × 6.16 × 60 × 60) C/4(96500) C
x g = 41.4 g
Learn more about electrolysis: brainly.com/question/12054569
<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
Answer:
2Cu2^+ + 2I^- ----> 2Cu^+ + I2
Explanation:
The reaction performed in the experiment is;
2 Cu(NO3)2 + 4 KI → 2 CuI (s) + 4 KNO3 + I2
The iodide ions reduces Cu^2+ to Cu^+ which is insoluble in water hence the precipitate. This is so because iodine is a good oxidizing agent seeing that it requires one electron to fill its outermost shell. Potassium on the other hand is a good reducing agent since it easily looses its one electron.
The oxidation - reduction equation is as follows;
2Cu2^+ + 2e ----> 2Cu^+ reduction half equation
2I^- ----> I2 + 2e. Oxidation half equation
Balanced redox reaction equation;
2Cu2^+ + 2I^- ----> 2Cu^+ + I2
Since we know that one mole of any gas at STP is equal to 22.4 L we can multiply 135L by the following conversion: 1 mole/22.4L. When you set up the problem it looks like this…: (135L)x 1 mole/22.4L =6.03 moles of oxygen gas The liters cancel out and you are left with moles as your units.
So your answer is then 3.058
Answer:
<span>ρ≅13.0⋅g⋅m<span>L<span>−1</span></span></span> = <span>13.0⋅g⋅c<span>m<span>−3</span></span></span>
Explanation:
<span>Density=<span>MassPer unit Volume</span></span> = <span><span>75.0⋅g</span><span><span>(36.5−31.4)</span>⋅mL</span></span> <span>=??g⋅m<span>L<span>−1</span></span></span>
Note that <span>1⋅mL</span> = <span>1⋅c<span>m<span>−3</span></span></span>; these are equivalent units of volume;
i.e. <span>1⋅c<span>m3</span></span> = <span>1×<span><span>(<span>10<span>−2</span></span>⋅m)</span>3</span>=1×<span>10<span>−6</span></span>⋅<span>m3</span>=<span>10<span>−3</span></span>⋅L=1⋅mL</span>.
