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Gnesinka [82]
3 years ago
9

The spontaneous reaction that occurs when the cell in the picture operates is as follows: 2Ag+ + Cd(s) ???? 2 Ag(s) + Cd2+ (A) V

oltage increases. (B) Voltage decreases but remains > zero. (C) Voltage becomes zero and remains at zero. (D) No change in voltage occurs. (E) Direction of voltage change cannot be predicted without additional information. Which of the above occurs for each of the following circumstances? 14. A 50-milliliter sample of a 2-molar Cd(NO3)2 solution is added to the left beaker. 15. The silver electrode is made larger. 16. The salt bridge is replaced by a platinum wire. 17. Current is allowed to flow for 5 minutes.
Chemistry
1 answer:
Snowcat [4.5K]3 years ago
7 0

Answer:

14. B    15. D    16. C     17. B

Explanation:

The spontaneous reaction that occurs when the cell operates is shown below:

2Ag^{+} + Cd_{(s)} ⇒2Ag_{(s)} + Cd^{2+}

We need to select the correct option from the list below for the following questions.

(A) Voltage increases. (B) Voltage decreases but remains > zero. (C) Voltage becomes zero and remains at zero. (D) No change in voltage occurs. (E) Direction of voltage change cannot be predicted without additional information.

14. A 50-milliliter sample of a 2-molar Cd(NO_{3})_{2} solution is added to the left beaker.

If a 50-milliliter sample of a 2-molar Cd(NO_{3})_{2}  solution is added to the left beaker, the voltage decreases but its value remains greater than zero. The correct option is B

15. The silver electrode is made larger.

If the silver electrode is made larger, no change in the value of the voltage since we don't have the idea of the initial value. The correct option is D.

16. The salt bridge is replaced by a platinum wire.

If the salt bridge is replaced by a platinum wire, there will be no passage of electrons because electrons can't pass through a platinum wire. Therefore, the voltage will be zero and remains at zero. The correct option is C.

17. Current is allowed to flow for 5 minutes.

If current is allowed to flow for 5 minutes, the voltage decreases but its value remains greater than zero. The correct option is B.

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Answer:

For n=3 and l=1=p

It is 3p-orbital.

Magnetic quantum number m

l

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Forl=1, m

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m

l

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Each orbital can occupy maximum of two electron

Number of electrons =2×3=6

Thus 6 electrons will show same quantum number values of n=3 and l=1.

Number of elements with last electron in 3p orbitals = 6

5 0
3 years ago
Which of the following compounds has the general formula X3Y2?Which of the following compounds has the general formula X3Y2?
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Explanation:

6 0
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Which of the following can be changed in a chemical reaction?
DanielleElmas [232]

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4 0
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Dinitrogen tetraoxide, N2O4, decomposes to nitrogen dioxide, NO2, in a first-order process. If k = 1.5 x 103 s-1 at 5 ºC and k =
Arada [10]

Answer:

The activation energy for the decomposition = 33813.28 J/mol

Explanation:

Using the expression,

\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )

Wherem  

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

E_a is the activation energy

R is Gas constant having value = 8.314 J / K mol  

Thus, given that, E_a = ?

k_2=4.0\times 10^3s^{-1}

k_1=1.5\times 10^3s^{-1}  

T_1=5\ ^0C  

T_2=25\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (5 + 273.15) K = 278.15 K  

T = (25 + 273.15) K = 298.15 K  

T_1=278.15\ K

T_2=298.15\ K

So,

\ln \frac{1.5\times 10^3}{4.0\times 10^3}\:=-\frac{E_{a}}{8.314}\times \left(\frac{1}{278.15}-\frac{1}{298.15}\right)

E_a=-\ln \frac{1.5\times \:10^3}{4.0\times \:10^3}\:\times \frac{8.314}{\left(\frac{1}{278.15}-\frac{1}{298.15}\right)}

E_a=-\frac{8.314\ln \left(\frac{1.5\times \:10^3}{4\times \:10^3}\right)}{\frac{1}{278.15}-\frac{1}{298.15}}

E_a=-\frac{689483.53266 \ln \left(\frac{1.5}{4}\right)}{20}

E_a=33813.28\ J/mol

<u>The activation energy for the decomposition = 33813.28 J/mol</u>

8 0
3 years ago
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