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densk [106]
3 years ago
14

Yet another reaction has an equilibrium constant kc=4.32×105 at 25 ∘c. it is an exothermic reaction, giving off quite a bit of h

eat while the reaction proceeds. if the temperature is raised to 200 ∘c , what will happen to the equilibrium constant? the equilibrium constant will yet another reaction has an equilibrium constant at 25 . it is an exothermic reaction, giving off quite a bit of heat while the reaction proceeds. if the temperature is raised to 200 , what will happen to the equilibrium constant?
Chemistry
2 answers:
Tatiana [17]3 years ago
8 0
The equilibrium constant will decrease.
when this reaction is an exothermic reaction, SO we can assume heat as a product of that reaction.

when Kc = concentration of product/concentration of reactants

So when the temperature and heat increase the reaction will go in the direction which decreases the heat and achieves equilibrium again. So it will go leftwards to decrease the products. and when the products decrease the Kc will decrease. 

Kitty [74]3 years ago
4 0
<span><u><em>Answer:</em></u>
The equilibrium constant will decrease

<u><em>Explanation:</em></u>
<u>The equilibriuc constant (Kc) can be calculated as follows:</u>
Kc = </span>\frac{concentration_o_fproducts}{concentration_o_freactants}<span>

Now, the given equation is an exothermic reaction. This means that <u>heat is s product.</u>

If we increase the amount of heat, the system will no longer be at equilibrium. This equilibrium will <u>shift to the left creating more reactants</u> in order to balance the system and be reach equilibrium again.

By this, we can note that the <u>concentration of reactants in the system will increase</u> which means that the equilibrium constant will decrease.

Hope this helps :)</span>
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The fermentation of C6H12O6 will produce carbon dioxide and
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Answer:

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Explanation:

CO2 is carbon dioxide

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Wewaii [24]

Answer:

turgor pressure can be done in a lab or a self test.

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Explanation:

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7 0
3 years ago
From the value Kf=1.2×109 for Ni(NH3)62+, calculate the concentration of NH3 required to just dissolve 0.016 mol of NiC2O4 (Ksp
Nina [5.8K]

<u>Answer:</u> The concentration of NH_3 required will be 0.285 M.

<u>Explanation:</u>

To calculate the molarity of NiC_2O_4, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles of NiC_2O_4 = 0.016 moles

Volume of solution = 1 L

Putting values in above equation, we get:

\text{Molarity of }NiC_2O_4=\frac{0.016mol}{1L}=0.016M

For the given chemical equations:

NiC_2O_4(s)\rightleftharpoons Ni^{2+}(aq.)+C_2O_4^{2-}(aq.);K_{sp}=4.0\times 10^{-10}

Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9

Net equation: NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?

To calculate the equilibrium constant, K for above equation, we get:

K=K_{sp}\times K_f\\K=(4.0\times 10^{-10})\times (1.2\times 10^9)=0.48

The expression for equilibrium constant of above equation is:

K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}

As, NiC_2O_4 is a solid, so its activity is taken as 1 and so for C_2O_4^{2-}

We are given:

[[Ni(NH_3)_6]^{2+}]=0.016M

Putting values in above equations, we get:

0.48=\frac{0.016}{[NH_3]^6}}

[NH_3]=0.285M

Hence, the concentration of NH_3 required will be 0.285 M.

7 0
3 years ago
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tiny-mole [99]

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8 0
2 years ago
A saturated solution is made by dissolving 0.327 g of a polypeptide (a substance formed by joining together in a chainlike fashi
faust18 [17]

Answer: The approximate molecular mass of the polypeptide is 856 g/mol

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

\pi=iMRT

Or,

\pi=i\times \frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution in L)}}\times RT

where,

\pi = osmotic pressure of the solution = 4.19 torr

i = Van't hoff factor = 1 (for non-electrolytes)

Mass of solute (polypeptide) = 0.327 g

Volume of solution = 1.70 L

R = Gas constant = 62.364\text{ L.torr }mol^{-1}K^{-1}

T = temperature of the solution = 26^oC=[273+26]K=299K

Putting values in above equation, we get:

4.19torr=1\times \frac{0.327}{\text{Molar mass of solute}\times 1.70}\times 62.364\text{ L.mmHg }mol^{-1}K^{-1}\times 299K\\\\\text{molar mass of solute}=856g/mol

Hence, the molar mass of the polypeptide is 856 g/mol

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3 years ago
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