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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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A system gains 652 kJ of heat, resulting in a change in internal energy of the system equal to +241 kJ. How much work is done?
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Answer:

-411 kj

Explanation:

We solve by using this formula

∆U = ∆Q + ∆W

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Putting the value into the equation

+241 = 652 + W

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Workdone = -411 kj

Since work done is negative it means that work was done by the system

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_NaHCO(s) --&gt; _CO2+_NaCO(s)+_H2O<br><br>balance the equation​
elena55 [62]

The balanced equation :

2NaHCO₃⇒CO₂ + Na₂CO₃+H₂O

<h3>Further explanation</h3>

Given

Reaction

NaHCO(s) --> _CO2+_NaCO(s)+_H2O

Required

The balanced equation

Solution

Maybe the equation should be like this :

NaHCO₃⇒CO₂ + Na₂CO₃+H₂O

Give a coefficient

NaHCO₃⇒aCO₂ + bNa₂CO₃+cH₂O

Make an equation

Na, left=1, right=2b⇒2b=1⇒b=1/2

H, left=1, right=2c⇒2c=1⇒c=1/2

C, left=1, right=a+b⇒a+b=1⇒a+1/2=1⇒a=1/2

The equation becomes :

NaHCO₃⇒1/2CO₂ +1/2Na₂CO₃+1/2H₂O x2

2NaHCO₃⇒CO₂ + Na₂CO₃+H₂O

5 0
2 years ago
Why is the air in a jet aircraft flying at high altitudes pressurized? *
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Answer:

Aircraft cabins are therefore pressurized to maintained a similar pressure as that experienced at sea level to ensure normal breathing of passengers.

Explanation:

-Air becomes increasingly thinner with increasing altitudes.

-As such, oxygen becomes limited at higher altitudes and makes it difficult or almost impossible to breath  a condition called hypoxia.

-Aircraft cabins are therefore pressurized to maintained a similar pressure as that experienced at sea level to ensure normal breathing of passengers.

6 0
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