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3 years ago

1.72 mol of LiCl in 37.5 L of solution

Chemistry

1 answer:

givi [52]3 years ago

4 0

<h3>Answer:</h3>

Molarity is 0.046 M

<h3>Explanation:</h3>

We are given;

1.72 mol of LiCl in 37.5 L of solution

We will take the question to be; calculate the molarity of LiCl

Therefore,

we can start by defining molarity as the concentration of a solution in moles per liter.

Molarity of a solution is calculated by dividing the number of moles of solute by the volume of solution.

Molarity = Moles of solute ÷ Volume of the solution

Thus, in this case;

Molarity of LiCl = Moles of LiCl ÷ Volume of the solution

= 1.72 moles ÷ 37.5 L

= 0.0459 M

= 0.046 M

Therefore, the molarity of LiCl solution is 0.046 M

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Answer:

From molar mass=total RAM of each individual element

78.8=(16+1)×3+M

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2 years ago

Question 1 (4 points)

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2 years ago

Combine the two half-reactions that give the spontaneous cell reaction with the smallest E∘. Fe2+(aq)+2e−→Fe(s) E∘=−0.45V I2(s)+

Iteru [2.4K]

<u>Answer:</u> The spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

<u>Explanation:</u>

We are given:

$E^o_{(Fe^{2+}/Fe)}=-0.45V\\E^o_{(I_2/I^-)}=0.54V\\E^o_{(Cu^{2+}/Cu)}=0.34V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, iodine will always undergo reduction reaction, then copper and then iron.

The equation used to calculate electrode potential of the cell is:

$E^o_{cell}=E^o_{oxidation}+E^o_{reduction}$

The combination of the cell reactions follows:

<u>Case 1:</u>

Here, iodine is getting reduced and iron is getting oxidized.

The cell equation follows:

$I_2(s)+Fe(s)\rightarrow Fe^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=0.45+0.54=0.99V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

<u>Case 2:</u>

Here, iodine is getting reduced and copper is getting oxidized.

The cell equation follows:

$I_2(s)+Cu(s)\rightarrow Cu^{2+}(aq.)+2I^-(aq.)$

Oxidation half reaction: $Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-$ $E^o_{oxidation}=-0.34V$

Reduction half reaction: $I_2(s)+2e^-\rightarrow 2I_-(aq.)$ $E^o_{reduction}=0.54V$

$E^o_{cell}=-0.34+0.54=0.20V$

Thus, this cell will give the spontaneous cell reaction with smallest $E^o_{cell}$

<u>Case 3:</u>

Here, copper is getting reduced and iron is getting oxidized.

The cell equation follows:

$Cu^{2+}(aq.)+Fe(s)\rightarrow Fe^{2+}(aq.)+Cu(s)$

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-$ $E^o_{oxidation}=0.45V$

Reduction half reaction: $Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)$ $E^o_{reduction}=0.34V$

$E^o_{cell}=0.45+0.34=0.79V$

Thus, this cell will not give the spontaneous cell reaction with smallest $E^o_{cell}$

Hence, the spontaneous cell reaction having smallest $E^o$ is $I_2+Cu\rightarrow Cu^{2+}+2I^-$

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3 years ago

When the vapor pressure of water is 0.106 mmHg, determine the reaction quotient for that above equilibrium.

TEA [102]

Answer:

Answer is explained in the explanation section below.

Explanation:

Note: This question is not complete and lacks necessary data to solve. However, I have found a similar question and I will be using its data to solve this question for the sake of understanding and concept.

Solution:

Equilibrium Reaction:

CaO(s) + H2O(g) -->Ca(OH)2(s)

We need to find the reaction quotient for this question:

Q = $\frac{1}{P_{H20} }$