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yulyashka [42]
4 years ago
6

H2+I2= 2HI+3 kcal Based on the reaction given above, as the temperature decreases, the formation of the reactants will: A. decre

ase B. increase C. remained consistent D. indeterminate
Chemistry
2 answers:
kramer4 years ago
4 0
Answer:
Decrease

Explanation:
The reaction given is an exothermic reaction. This means that the heat is a product.
At equilibrium, the rate of forward reaction is equal to the rate of backward reaction.
Now, decreasing the temperature means that the concentration of the products will decrease. 
As a result, the equilibrium will shift to the right leading to the formation of more products in order to reach equilibrium again.
This means that the formation of the reactants will decrease.

Hope this helps :)
In-s [12.5K]4 years ago
4 0

the answer is A: decrease

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nordsb [41]

Answer:

C.) 3.5 atm

Explanation:

Because you are dealing with pressure and volume, you can use Boyle's Law to help you find the missing variable. The general formula looks like this:

P₁V₁ = P₂V₂

In this formula, "P₁" and "V₁" represent the pressure and volume of the chamber originally. "P₂" and "V₂" represent the final pressure and volume. You have been given the values of all the other variables but "P₂". Therefore, you can plug these values into the formula and simplify to find the final pressure.

P₁ = 1.5 atm

V₁ = 1.4 L

P₂ = ?

V₂ = 0.60 L

P₁V₁ = P₂V₂                                        <---- Boyle's Law formula

(1.5 atm)(1.4 L) = P₂(0.60 L)               <---- Plug in values

2.1 = P₂(0.60 L)                                 <---- Multiply 1.5 and 1.4

3.5 = P₂                                             <----- Divide both sides by 0.60

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2 years ago
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What is the new freezing point for a 0.811m solution of Li2O in water?
rewona [7]

Answer:

Freezing T° of solution = - 4.52°C

Explanation:

ΔT = Kf . m . i

That's the formula for colligative property about freezing point depression.

Li₂O is an oxide that can not be dissociated but, if we see it's a ionic compound.

Li₂O →  2Li⁺  +  O⁻²

3 moles of ions have been formed. Ions dissolved in solution are i, what we call Van't Hoff factor.

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We replace: 0°C - Freezing T° of solution = 1.86°C/m . 0.811 m . 3

Freezing T° of solution = - 4.52°C

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