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fredd [130]
3 years ago
12

What is the value of the equilibrium constant for this redox reaction? 2Cr3+(aq) + 3Sn2+ (aq) 2Cr(s) + 3Sn4+(aq) E = -0.89 v

Chemistry
2 answers:
charle [14.2K]3 years ago
8 0
<span>b. square root of 3 over 3 is the answer to your question!!! I hope I helped!!!!!!!!!!                                                                 XoXo -Marcey<3!  :D
</span>
Kay [80]3 years ago
6 0

Answer : The correct option is, (B) K=6.27\times 10^{-91}

Solution :

The given balanced redox reaction is,

2Cr^{3+}(aq)+3Sn^{2+}(aq)\rightarrow 2Cr(s)+3Sn^{4+}(aq)

Now we have to calculate the equilibrium constant for the redox reaction.

The relation between the equilibrium constant and cell potential :

\Delta G=-nFE^o_{cell}\\\\\Delta G=-2.303RT\log K

By equation these two equation we get,

E^o_{cell}=\frac{0.0592}{n}\times \log K       ........(1)

where,

E^o_{cell} = cell potential = -0.89 v

n = number of electrons = 6

K = equilibrium constant

Now put all the given values in equation (1), we get

-0.89v=\frac{0.0592}{6}\times \log K

\log K=-90.202

K=6.28\times 10^{-91}

Therefore, the value of the equilibrium constant is, K=6.27\times 10^{-91}

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How many moles are in 4.3 x 10 ^22 molecules of Na3PO4
Sever21 [200]

Answer:

0.071 moles of Na₃PO₄ .

Explanation:

Given data:

Number of molecules of Na₃PO₄ = 4.3× 10²² molecules

Number of moles = ?

Solution:

1 mole contain 6.022 × 10²³ molecules

4.3× 10²² molecules × 1 mol / 6.022 × 10²³ molecules

0.71× 10⁻¹ mol

0.071 mol

The number 6.022 × 10²³ is called Avogadro number.

"It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance"

3 0
3 years ago
What is the name of the gas produced when nitric acid is added to copper metal? please help! much appreciated!
Lemur [1.5K]
The answer to your question is nitrogen dioxide

8 0
3 years ago
Least to greatest 2.62,2 2/5, 26.8%, 2.26%,271%
olga_2 [115]
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Hope this helps! :D
3 0
3 years ago
Read 2 more answers
The chart shows the solubility of different substances.
Annette [7]
Answer: 1) Temperature can change the solubility of a solute.

Explanation:

The chart is missing so there is no way to tell what does the graph show.

Yet, I can help you because I can explain the status of each statement of the choices. As you will see there is only one possibility..

<span>1) Temperature can change the solubility of a solute.

Yes, temperature definetly can, and mostly do, modify the solubility of a solute.

You can search any chart of solubility and will find that.

I can give you two examples:

a) Sodium chloride: dissolve some spoons of salt in a cold water  until you can not dissolve more. Then, heat the water, you will find that more salt will get dissolved, proving that the temperature of the solution increases the solubility of sodium chloride.

b) Carbon dioxide gas: the soft drinks have CO₂ molecules dissolved in it.
 
The higher the temperature of the soft drink the less the amount of CO₂(g) that can be dissolved. That is why the soda bottling plants cool the beverage before adding the CO₂(g).

2) </span><span>Temperature has no affect on the solubility of a solute.

Since this is the opposite to the first statement and the first is true, this is false.

3) Salt has a greater solubility than sugar.

False.

This is an empirical result, which you cannot predict theoretically. So you need to see at the data either in a table or in a chart. Else you can test it at home. After the empirical data are shown it results that more grams of sugar can be dissolved in water compared to salt.

That is something you ca see in a chart or you can prove by yourself.

4) Nitrite salt has a greater solubility than sugar. </span>

False.

Looking at some data you can find that sodium nitrite solutiliby is aroun  70 - 100 g/10 g while sugar (sucrose) solutiblity is around 180 - 235 g/ 100 g.

8 0
3 years ago
Read 2 more answers
Water vapor enters a compressor at 35 kpa and 160°c and leaves at 300 kpa with the same specific entropy as at the inlet. What a
Annette [7]
<h3><em><u>solution</u></em><em><u>:</u></em></h3>

<em><u>The initial entropy is obtained from the initial pressure and temperature with data from A-6 using interpolation:</u></em>

<em><u>s</u></em><em><u>=</u></em><em><u> </u></em><em><u>8</u></em><em><u>.</u></em><em><u>26</u></em><em><u>5</u></em><em><u>2</u></em><em><u> </u></em><em><u>kJ</u></em><em><u>/</u></em><em><u>kgK</u></em>

<em><u>The final temperature is determined from the entropy and the final pressure with data from A-6 using interpolation:</u></em>

<em><u>T₂ = T₁+</u></em><em><u> </u></em><em><u>T₂ - </u></em><em><u>T₁</u></em><em><u>/</u></em><em><u> </u></em><em><u>8</u></em><em><u>₂</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>8</u></em><em><u>₁</u></em><em><u> </u></em><em><u>(</u></em><em><u> </u></em><em><u>s</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>s</u></em><em><u>₁</u></em><em><u>)</u></em>

<em><u>= </u></em><em><u>(</u></em><em><u>400 +</u></em><em><u> </u></em><em><u>500 - 400</u></em><em><u>/</u></em><em><u>8.3271</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>8.0347</u></em><em><u> </u></em><em><u>(8.2652 - 8</u></em><em><u>)</u></em><em><u>)</u></em>

<em><u>= 478.83°C</u></em>

<em><u>The final enthalpy is determined in the same way:</u></em>

<em><u>h₂= h₁</u></em><em><u> </u></em><em><u>+</u></em><em><u> </u></em><em><u>h₂</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>h₁</u></em><em><u>/</u></em><em><u>s</u></em><em><u>₂</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>s</u></em><em><u>₁</u></em><em><u> </u></em><em><u>( s - s₁)</u></em>

<em><u>= (</u></em><em><u> </u></em><em><u>3275.5</u></em><em><u>+</u></em><em><u> </u></em><em><u>3486.6 </u></em><em><u>-</u></em><em><u> </u></em><em><u>3275.5</u></em><em><u>/</u></em><em><u> </u></em><em><u>8.3271</u></em><em><u> </u></em><em><u>-</u></em><em><u> </u></em><em><u>8.0347</u></em><em><u>)</u></em><em><u> </u></em><em><u>(8.265</u></em><em><u>)</u></em>

<em><u>=</u></em><em><u> </u></em><em><u>3441.91 </u></em><em><u>kJ</u></em><em><u>/</u></em><em><u>kg</u></em>

6 0
2 years ago
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