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2 years ago

A solution of carbonic acid is at equilibrium. How would the system change is more carbonic acid was added to the solution

Chemistry

1 answer:

7nadin3 [17]2 years ago

3 0

Answer:

The equilibrium position shifts to the right, in accordance to the constraint principle

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Predict the products of the reaction below. That is, complete the right-hand side of the chemical equation. Be sure your equatio

Furkat [3]

Answer:

HNO₃ (aq) —> H⁺ (aq) + NO₃¯ (aq)

Explanation:

From the question given above

HNO₃ + H₂O —> ?

Nitric acid, HNO₃ reacts with water, H₂O to form aqueous solution of nitric acid as illustrated below:

HNO₃ + H₂O —> HNO₃ (aq)

Nitric acid is a strong acid and, so will ionised completely when dissolved in water. This is illustrated below:

HNO₃ (aq) —> H⁺ (aq) + NO₃¯ (aq)

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3 years ago

What reaction conditions most effectively conver a cabocxylic acid to a methly ester?

VLD [36.1K]

Answer:

Esterification reaction

Explanation:

When we have to go from an acid to an ester we can use the <u>esterification reaction</u>. On this reaction, an alcohol reacts with a carboxylic acid on acid medium to produce an ester and water. (See figure).

In this case, we need the <u>methyl ester</u>, therefore we have to choose the <u>appropriate alcohol</u>, so we have to use the <u>methanol</u> as reactive if we have to produce the methyl ester.

7 0

2 years ago

HC2H3O2 (aq) + H2O (l) ⇔ C2H3O2- (aq) + H3O+ (aq) Ka = 1.8 x 10-5

marin [14]

The concentration of [H3O⁺]=2.86 x 10⁻⁶ M

<h3>Further explanation</h3>

In general, the weak acid ionization reaction

HA (aq) ---> H⁺ (aq) + A⁻ (aq)

Ka's value

$\large {\boxed {\bold {Ka \: = \: \frac {[H ^ +] [A ^ -]} {[HA]}}}}$

Reaction

HC₂H₃O₂ (aq) + H₂O (l) ⇔ (aq) + H₃O⁺ (aq) Ka = 1.8 x 10⁻⁵

$\tt Ka=\dfrac{[C_2H_3O^{2-}[H_3O^+]]}{[HC_2H_3O_2]}}\\\\1.8\times 10^{-5}=\dfrac{0.22\times [H_3O^+]}{0.035}$

[H₃O⁺]=2.86 x 10⁻⁶ M

5 0

3 years ago

The chemical equation for the reaction of baking soda (sodium bicarbonate, nahco3) and vinegar (acetic acid, ch3cooh) may be wri

satela [25.4K]

<h3><u>Answer;</u></h3>

Step 1; NaHCO3(s) + CH3COOH(l)

Step 2 ; CO2(g)

<h3><u>Explanation;</u></h3>

The chemical equation for the reaction of baking soda (sodium bicarbonate, NaHCO3) and vinegar (acetic acid, CH3COOH) reaction occurs in two steps.

Step 1;

A double displacement reaction in which acetic acid in the vinegar reacts with sodium bicarbonate to form sodium acetate and carbonic acid:
Equation;

NaHCO3(s)+ CH3COOH(l) → CH3COONa(aq) + H2CO3(l)

Step 2;

Carbonic acid is unstable and undergoes a decomposition reaction to produce the carbon dioxide gas:

H2CO3(l) → H2O(l) + CO2(g)

3 0

2 years ago

Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th

mario62 [17]

<u>Answer:</u> The freezing point of solution is 2.6°C

<u>Explanation:</u>

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

$\Delta T_f$ = $\text{Freezing point of pure solution}-\text{Freezing point of solution}$

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point depression constant = 5.12 K/m = 5.12 °C/m

$m_{solute}$ = Given mass of solute (anthracene) = 7.99 g

$M_{solute}$ = Molar mass of solute (anthracene) = 178.23 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC$

Hence, the freezing point of solution is 2.6°C

8 0

2 years ago