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Artist 52 [7]
3 years ago
5

A 25.00 mL sample of the ammonia solution

Chemistry
1 answer:
musickatia [10]3 years ago
3 0

Answer:

1.634 molL-1

Explanation:

The mol ration between NH3 and HCl is 1 : 1

Using Ca Va / Cb Vb = Na / Nb   where a = acid and b = base

Na = 1

Nb = 1

Ca = 0.208 molL-1

Cb = ?

Va = 19.64 mL

Vb = 25.00mL

Solving for Cb

Cb = Ca Va / Vb

Cb = 0.208 * 19.64 / 25.0

Cb = 0.1634 molL-1 (Concentration of diluted ammonia solution)

Using the dilution equation;

C1V1 = C2V2

Initial Concentration, C1 = ?

Initial Volume, V1 = 25.00 mL

Final Volume, V2 =  250 mL

Final Concentration, C2 = 0.1634 molL-1

Solving for C1;

C1 = C2 * V2 / V1

C1 = 0.1634 * 250 / 25.00

C1 = 1.634 molL-1

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How do you represent a shared pair of electrons when drawing the Lewis structure of a covalent bond
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6 0
3 years ago
Fill in the left side of this equilibrium constant equation for the reaction of benzoic acid (HC_6H_5CO_2) with water.
il63 [147K]

Answer:

[C₆H₅COO⁻][H₃O⁺]/[C₆H₅COOH] = Ka

Explanation:

The reaction of dissociation of the benzoic acid in water is given by the following equation:

C₆H₅-COOH + H₂O  ⇄  C₆H₅-COO⁻  +  H₃O⁺    (1)

The dissociation constant of an acid is the measure of the strength of an acid:

HA ⇄ A⁻ + H⁺        (2)

K_{a} = \frac{[A^{-}][H^{+}]}{[HA]}      (3)

<em>Where the dissociation constant of the acid (Ka) is equal to the ratio of the concentration of the dissociated forms of the acid, [A⁻][H⁺], and the concentration of the acid, [HA].     </em>

So, starting from the equations (2) and (3), the constant equation for the dissociation reaction of benzoic acid in water, of the equation (1), is:

K_{a} = \frac{[C_{6}H_{5}COO^{-}][H_{3}O^{+}]}{[C_{6}H_{5}COOH]}

I hope it helps you!          

6 0
3 years ago
What is the percentage of lithium in lithium carbonate (Li2CO3)?
ale4655 [162]
Molar mass Li2CO3 = 73.89 g/mol
Molar mass Li = 6.94g/mol Li = 6.94*2 = 13.88g


% LI = 13.88/73.89*100 = 18.78% perfectly correct.
6 0
3 years ago
Read 2 more answers
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