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2 years ago

What are the two species in highest concentration in a 0.1 M solution of lactic acid (CH3CH(OH)C00H), (Ka

Chemistry

1 answer:

rusak2 [61]2 years ago

6 0

Answer:

1.4 × 10^-4.

Explanation:

C3H6O3 + H2O <======> C3H5O3^- + H3O^+ ------------------------------------------(1).

So, from the question above we are given the following parameters or data which is going to help in solving this particular Question/problem;

=>concentration of the solution of lactic acid (CH3CH(OH)C00H) = 0.1 M and pH = 2.44.

Therefore, the concentration of the hydrogen ion[H^+} can be determined from the pH formula given below;

pH = - log { H^+}.

2.44 = - log { H^+}.

Therefore, {H^+} = 0.0036 M.

From the equation (1) given above, we have that the ratio for the equilibrium reaction is 1 : 1 : 1 :1. Therefore, molarity of C3H5O3^- = 0.0036 M and the molarity of C3H6O3 =( 0.1 - 0.0036 M) = 0.0964 M at equilibrium.

Hence, ka = {C3H5O3^-} { H3O^+} /{C3H6O3} = ( 0.0036 M)^2 /(0.0964 M) = 1.4 × 10^-4.

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3 years ago

Which substance is the oxidizing agent in the following reaction? Fe2S3 + 12HNO3 → 2Fe(NO3)3 + 3S + 6NO2 + 6H2O Which substance

aleksandr82 [10.1K]

Answer:

HNO3 is the oxidizing agent

Explanation:

Step 1:

The oxidizing agent is responsible to oxidize another. but itself undergoes a reduction .

Fe2S3:

Fe has an oxidation number of +3

S has an oxidation number of -2

HNO3:

O has an oxidation number of -2 ( we have 3 times O so this makes -6)

H has an oxidation number of +1

N has an oxidation number of +5

Fe(NO3)3

O has an oxidation number of -2 ( we have 3 times O so this makes -6)

Fe has an oxidation number of +3

Since NO3 has an oxidation number of -1; N has an oxidation number of +5

S alone has an oxidation number of 0

NO2:

O has an oxidation number of -2 (we have 2 times O, this makes -4)

N has an oxidation number of +4

Fe doesn't change from oxidation number. It stays +3

N goes from +5 to +4 → this is a reduction

S goes from -2 to 0 → this is an oxidation

The reducing agent is the compound that contributes the oxidized species (S).

The oxidizing agent contributes the reducing species (N)

The answer is HNO3

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2 years ago

5.0 gram of anhydrous calcium carbonate is reacted with an excess of dilute nitric acid. if the reaction is carried out at 25 de

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Answer:

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Explanation:

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2 years ago

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Answer:

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Explanation:

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2 years ago

Read 2 more answers

1.12g H2 is allowed to react with 9.60 g N2, producing 1.23 g NH3.

andriy [413]

Answer:

A. $m_{NH_3}^{theo} =1.50gNH_3$

B. $Y=82.2\%$

Explanation:

Hello!

In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:

$N_2+3H_2\rightarrow 2NH_3$

Thus we proceed as follows:

A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:

$n_{NH_3}^{by \ H_2}=1.12gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.370molNH_3\\\\ n_{NH_3}^{by \ N_2}=1.23gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.0878molNH_3$

Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:

$m_{NH_3}^{theo}=0.0878mol*\frac{17.04gNH_3}{1molNH_3} =1.50gNH_3$

B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:

$Y=\frac{1.23gNH_3}{1.50gNH_3} *100\%\\\\Y=82.2\%$

Best regards!

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2 years ago