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Calculate the pH of the cathode compartment solution if the cell emf at 298 K is measured to be 0.660 V when (Zn^2+)=0.22 M and(P_H2)= 0.87atm.

Answers

Answer 1

Answer:

Answer:

pH = 2.059

Explanation:

At the Cathode:

The reduction reaction is:

$2H^+ + 2e^- \to H_2 \ \ \ \mathbf{E^0_(red)= 0.00 \ V}$

At the anode:

At oxidation reaction is:

$Zn \to Zn^(2+) +2e^- \ \ \ \mathbf{E^0_(ox) = 0.76 \ V}$

The overall equation for the reaction is:

$\mathbf{Zn + 2H^+ \to Zn^(2+) + H_2}$

The overall cell potential is:

$\mathbf{E^0_(cell)= E^0_(ox) + E^0_(red)}$

$\mathbf{E^0_(cell)= 0.76 \ V +0.00 \ V}$

$\mathbf{E^0_(cell)= 0.76\ V}$

Using the formula for the Nernst equation:

$E = E^0 - ( (0.0591)/(n))log (Q)\n$

where;

E = 0.66

(Zn^2+)=0.22 M

Then

$0.66 =0.76- ( (0.0591)/(2))log \bigg ( ([Zn^(2+) ] PH_2)/([H^+]^2) \bigg )$

$0.66 =0.76- 0.02955 * log \bigg ( (0.22*0.87)/([H^+]^2) \bigg )$

3.4 = log ( 0.1914) - 2 log [H⁺]

3.4 = -0.7180 - 2 log [H⁺]

3.4 + 0.7180 = - 2 log [H⁺]

4.118 = - 2 log [H⁺]

pH = log [H⁺] = 4.118/2

pH = 2.059

Answer 2

Answer:

The pH of the solution as described in the question is 2.7.

The equation of the reaction is;

Zn(s) + 2H^+(aq) ----> Zn^2+(aq) + H2(g)

The partial pressure of hydrogen can be converted to molarity using;

P= MRT

M = P/RT

M = 0.87atm/0.082 LatmK-1mol-1 × 298 K = 0.036 mol/L

We have to obtain the reaction quotient

Q = [Zn^2+] [H2]/[H^+]^2

Q = [0.22 ] [0.036]/[H^+]^2

Recall that, from Nernst equation;

E = E° - 0.0592/nlog Q

E° = 0.00V - (-0.76V) = 0.76V

0.660 = 0.76 - 0.0592/2logQ

0.660 - 0.76 = - 0.0592/2logQ

-0.1 = - 0.0592/2logQ

-0.1 × 2/ - 0.0592 = logQ

3.38 = log Q

Q = Antilog (3.38)

Q= 2.39 × 10^3

Now;

2.39 × 10^3 = [0.22 ] [0.036]/[H^+]^2

2.39 × 10^3 = 7.92 × 10^-3/[H^+]^2

[H^+]^2 = 7.92 × 10^-3/2.39 × 10^3

[H^+] = 1.82 × 10^-3

pH = -log[H^+]

pH = -log[ 1.82 × 10^-3]

pH = 2.7

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How many atoms are in 2.3 moles Au?

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Answer:

6.02×10^23atoms of au

Answer:

i think 3

Explanation:

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Answers

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You can get better at solving problems like this by practicing a lot!

Estimate how much heat in joules is released when 25.0 g of water (C = 4.184 J/g°C) is cooled from 80.0°C to 30.0°C?

Answers

The amount of heat will be 5230 j.

What is heat?

Heat is a type of energy that is transferred between both the system and its surroundings as a result of temperature variations.

Calculation of heat.

Given data:

Mass = 25.0 g = 0.025 kg

C = 4.184 J/g°C

$T_(1)$ = 80.0°C

$T_(2)$ = 30.0°C

Q= ?

By using the formula of heat.

Q = MC ( $T_(2) - T_(1)$ )

Put the value of given data in heat equation.

Q(heat) = 0.025 × 4.184 ( 30 - 80)

Q(heat) = 5230 J.

Therefore, the amount of heat will be 5230 J.

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Answer:

5230 J

Explanation:

m = 25 g = 0,025 kg

c = 4,184 J /(g * °C) = 4184 J /(kg * °C)

$t_(1)$ = 80 °C

$t_(2)$ = 30 °C

The formula is Q = c *m * ( $t_(2) - t_(1)$ )

Calculating:

Q = 4184 * 0,025 * (30 - 80) = 5230 J

Note that we get a negative heat (-5230 J). It just means that it is released.

Write the equilibrium expression, calculate KEQ and then tell where the equilibrium lies: Fe (s) + O2 (g) ↔ Fe2O3 (s) In a 2.0 L Container At equilibrium: Fe = 1.0 mol O2 = 1.0 E-3 mol Fe2O3 = 2.0 mol

Answers

Answer:

The value of Keq is 4e-9. See the solution below

Explanation:

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What is the toto number of neutrons in an atom of 73 li

Answers

The 73 in Li means that there are 73 protons and neutrons in this atom. To find the number of neutrons, simply subtract the number of protons, which is equal to the atomic number (which is 3 in this case).

70 neutrons.

Rank the following compounds in order of decreasing acid strength using periodic trends. Rank the acids from strongest to weakest. To rank items as equivalent, overlap them. H2Se

HBr

H2O

HI

Answers

Explanation:

It is known that acidic strength of hydrides of same group tends to increase when we move from top to bottom in a group. On the other hand, acidic strength of hydrides of same period elements increases when we move from left to right in a period.

As both bromine and iodine belongs to the same group. Also, selenium and oxygen are same group elements. Therefore, their acidic strength increases on moving down the group.

Therefore, we can conclude that acidic strength of given compounds from strongest to weakest is as follows.

HI > HBr > $H_(2)Se$ > $H_(2)O$

Final answer:

To rank the acids in decreasing acid strength using periodic trends, consider the size, electronegativity, and presence of lone pairs of electrons. HI is the strongest acid, followed by HBr, H2O, and H2Se.

Explanation:

To rank the acids in order of decreasing acid strength using periodic trends, we need to consider the size and electronegativity of the atoms. The larger the atom, the weaker the acid, and the more electronegative the atom, the stronger the acid. Additionally, we can consider the presence of lone pairs of electrons, as they increase the acidity.

HI - Iodine (I) is larger and less electronegative than the other halogens. It also has a lone pair of electrons, making it the strongest acid.
HBr - Bromine (Br) is larger and less electronegative than chlorine (Cl), and it also has a lone pair of electrons. It is the second strongest acid.
H2O - Oxygen (O) is smaller and more electronegative than the halogens. It does not have a lone pair of electrons, making it a weaker acid than the halogens.
H2Se - Selenium (Se) is larger and less electronegative than sulfur (S). However, it does not have a lone pair of electrons, making it the weakest acid.

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