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olya-2409 [2.1K]
3 years ago
7

What would cause the equilibrium to shift left in this reaction? CO + 3H2 ? CH4 + H2O A. Adding heat to the product mixture B. A

faster rate of forward reaction C. Increased collisions between CO and H2 D. The escape of water from the mixture E. Placing the mixture in a cold water bath
Chemistry
2 answers:
Irina-Kira [14]3 years ago
8 0

Answer:

Pretty sure the answer is A on Plato

Explanation:

I looked it up, and it said that raising the temperature would cause the equilibrium to shift left.

storchak [24]3 years ago
7 0

Answer:

Option A.

Explanation:

To decrease pressure by increasing volume, the equilibrium of the reaction shift to the left as the reactant side has greater number of moles than the product side.

Equilibrium also shifts to the left if temperature decreases.

Given equation is CO+3H_2\rightarrow CH_4+H_2O

In this case, equilibrium shifts to the left on <u>adding heat to the product mixture</u> .

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Many homeowners treat their lawns with CaCO3(s) to reduce the acidity of the soil. Write a net ionic equation for the reaction o
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The reaction of acid, assuming HCl and calcium carbonate always produces a gas. The reaction is as follows:
2 HCl + CaCO3 --> CaCl2 + H2CO3 
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H2CO3 --> H2O + CO2 
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2 HCl + CaCO3 --> CaCl2 + H2O + CO2 
The total ionic equation looks as follows: 
2H+(aq) + 2 Cl-(aq) + CaCO3(s) --> Ca+2(aq) + 2 Cl-(aq) + H2O(l) + CO2(g) 
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4 0
3 years ago
Submit What is the solubility of Cd3(POA) 2 in water? (Ksp of Cd3(PO4)2 is 2.5 x 10-33) | 1 2 3 +/- . 0 x100
vesna_86 [32]

<u>Answer:</u> The solubility of Cd_3(PO_4)_2 in water is 1.18\times 10^{-7}mol/L

<u>Explanation:</u>

The balanced equilibrium reaction for the ionization of cadmium phosphate follows:

Cd_3(PO_4)_2\rightleftharpoons 3Cd^{2+}+2PO_4^{3-}

                      3s       2s

The expression for solubility constant for this reaction will be:

K_{sp}=[Cd^{2+}]^3[PO_4^{3-}]^2

We are given:

K_{sp}=2.5\times 10^{-33}

Putting values in above equation, we get:

2.5\times 10^{-33}=(3s)^3\times (2s)^2\\\\2.5\times 10^{-33}=108s^5\\\\s=1.18\times 10^{-7}mol/L

Hence, the solubility of Cd_3(PO_4)_2 in water is 1.18\times 10^{-7}mol/L

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3 years ago
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