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Bumek [7]
3 years ago
12

Calculate the moles of calcium oxalate formed from 5 grams of potassium oxalate (molar mass 184.24 g/mol) the mole mole ratio fr

om the balanced equation is 2.1 potassium oxalate to calcium oxalate
Chemistry
1 answer:
Nataly_w [17]3 years ago
7 0

5 g of potassium oxalate react to produce 0.03 moles of calcium oxalate.

Calcium oxalate (CaC₂O₄) is obtained by the reaction of 5 g of potassium oxalate (K₂C₂O₄).

We can calculate the moles of CaC₂O₄ obtained considering the following relationships.

  • The molar mass of K₂C₂O₄ is 184.24 g/mol.
  • The mole ratio of K₂C₂O₄ to CaC₂O₄ is 2:1.

5 g K_2C_2O_4 \times \frac{1molK_2C_2O_4}{184.24gK_2C_2O_4}  \times \frac{1molCaC_2O_4}{2molK_2C_2O_4} = 0.03molCaC_2O_4

5 g of potassium oxalate react to produce 0.03 moles of calcium oxalate.

Learn more: brainly.com/question/15288923

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If a system has a reaction quotient of 2.13 ✕ 10−15 at 100°C, what will happen to the concentrations of COBr2, CO, and Br2 as th
qaws [65]

This is an incomplete question, here is a complete question.

Consider the following equilibrium at 100°C.

COBr_2(g)\rightleftharpoons CO(g)+Br_2(g)

K_c=4.74\times 10^4

Concentration at equilibrium:

[COBr_2]=1.58\times 10^{-6}M

[Co]=2.78\times 10^{-3}M

[Br_2]=2.51\times 10^{-5}M

If a system has a reaction quotient of 2.13 × 10⁻¹⁵ at 100°c, what will happen to the concentrations of COBr₂, Co and Br₂ as the reaction proceeds to equilibrium?

Answer : The concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.

Explanation :

Reaction quotient (Q) : It is defined as the measurement of the relative amounts of products and reactants present during a reaction at a particular time.

The given balanced chemical reaction is,

COBr_2(g)\rightleftharpoons CO(g)+Br_2(g)

The expression for reaction quotient will be :

Q=\frac{[CO][Br_2]}{[COBr_2]}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

Q=\frac{(2.78\times 10^{-3})\times (2.51\times 10^{-5})}{(1.58\times 10^{-6})}=4.42\times 10^{-2}

The given equilibrium constant value is, K_c=4.74\times 10^4

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

There are 3 conditions:

When Q>K_c that means product > reactant. So, the reaction is reactant favored.

When Q that means reactant > product. So, the reaction is product favored.

When Q=K_c that means product = reactant. So, the reaction is in equilibrium.

From the above we conclude that, the Q that means product < reactant. So, the reaction is product favored that means reaction must shift to the product (right) to be in equilibrium.

Hence, the concentrations of Co and Br₂ decreases and the concentrations of COBr₂ increases.

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3 years ago
What fraction of the original atoms of the radioactive sample will remain after the given number of half-lives has passed?
meriva

Answer:

1/8

Explanation:

\frac{1}{ {2}^{3} }  =  \frac{1}{8}

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