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lys-0071 [83]
3 years ago
6

Write the complete balanced equation for the reaction that occurs when calcium carbonate (CaCO3) reacts with lithium nitride (Li

3N) in an aqueous solution.
Chemistry
2 answers:
MrMuchimi3 years ago
6 0
For the reactants, we have:
calcium carbonate CaCO3
lithium nitride Li3N

For the products, we have:
lithium carbonate Li2CO3
calcium nitride Ca3N2

So, the basic UNBALANCED equation is:
CaCO3 + Li3N ................> Li2CO3 + Ca3N2

Now, we need to balance the equation. This means that we need the number of moles in reactants for each element to be equal to the number of moles in the product for this element.

Balancing the equation, we will get:
<span>3 CaCO3+ 2 Li3N .................> 3 Li2CO3+ Ca3N2</span>

Ann [662]3 years ago
4 0

Answer: 3CaCO_3(aq)+2Li_3N(aq)\rightarrow Ca_3N_2+3Li_2CO_3

Explanation: According to the law of conservation of mass, mass can neither be created nor be destroyed.

The mass of products must be same as that of the mass of reactants. For this the number of atoms of each element must be same on both sides of the arrow. Thus we need to balance the chemical equations.

3CaCO_3(aq)+2Li_3N(aq)\rightarrow Ca_3N_2+3Li_2CO_3

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Asexual reproduction involves the production of offspring without the need for a secondary partner to contribute genetic information.

Pollination and mating are two forms of sexual reproduction, so B, C, and D are incorrect. A, because the new potato grows directly from the old without the contribution of new genetic material, asexually reproduces.
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A 0.100 M solution of chloroacetic acid 1ClCH2COOH2 is 11.0% ionized. Using this information, calculate 3ClCH2COO-4, 3H 4, 3ClCH
bearhunter [10]

Answer:

[CICH2COOH] = 0.089 M

[CICH2COO-] = 0.011 M

[H+] = 0.011 M

Ka = 1.36 x 10^-3

Explanation:

The reaction

CICH2COOH  ⇄ H+ (aq) + CICH2COO- (aq)

The initial concentration of CICH2COOH is 0.10 M , chloroacetic acid is 11.0% ionized.

let us calculate Ka

First , find change in concentration

since , 11% ionized

change in concentration = 0.10 X 11% = 0.011 M

Initial Concentration of CICH2COOH = 0.10 M

change in concentration of CICH2COOH = - 0.011 M

Equilibrium Concentration of CICH2COOH = 0.10 M - 0.011 M = 0.089 m

Initial Concentration of CICH2COO- = 0 M

change in concentration of CICH2COO- = + 0.011 M

Equilibrium Concentration of CICH2COO- = 0.011 M

Initial Concentration of H+ = 0 M

change in concentration of H+ = + 0.011 M

Equilibrium Concentration of H+ = 0.011 M

Therefore,

[CICH2COOH] = 0.089 M

[CICH2COO-] = 0.011 M

[H+] = 0.011 M

Ka = [H+][CICH2COO-] /[CICH2COOH]

Ka = (0.011 * 0.011) / (0.089)

Ka = 1.36 x 10^-3

6 0
3 years ago
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