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rjkz [21]
3 years ago
9

At which point was neutralization complete for the acid that is being titrated below?

Chemistry
2 answers:
Hatshy [7]3 years ago
8 0

Answer:

D.

Explanation:

  • From the shown titration curve, It is a titration curve for a diprotic acid (H₂A).
  • The neutralization of the first proton is at point A.
  • The neutralization of the second proton (complete neutralization) is at point D.

<em>So, the right choice is: D.</em>

Rom4ik [11]3 years ago
8 0

D on e2020

extra words to get to 20 letters

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What is the oxidation state of CI in CIO-<br><br> -1<br> +2<br> +1<br> 0
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Draw a mechanism for the reaction of methylamine with 2-methylpropanoic acid. Draw any necessary curved arrows. Show the product
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Answer:

See figure 1

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5 0
3 years ago
For each of the following balanced chemical equations, calculate how many moles and how many grams of each product would be prod
natima [27]

Answer:

(a)

Moles of ammonium chloride = 0.5 mole

Mass of ammonium chloride formed = 26.7455 g

(b)

Mole of CS_2 = 0.125 mole

Mass = Moles * Molar mass = 0.125 * 76.139 g = 9.52 g

Mole of H_2S = 0.25 mole

Mass = Moles * Molar mass = 0.25 * 34.1 g = 8.525 g

Explanation:

(a)

For the first reaction:-

NH_3_{(g)}+HCl_{(g)}\rightarrow NH_4Cl_{(s)}

The mole ratio of the reactants = 1 : 1

0.5 moles of ammonia react with 0.5 moles of hydrochloric gas to give 0.5 moles of ammonium chloride

So, <u>Moles of ammonium chloride formed = 0.5 moles</u>

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<u>Mass = Moles * Molar mass = 0.5 * 53.491 g = 26.7455 g</u>

(b)

For the first reaction:-

CH_4_{(g)}+4S_{(s)}\rightarrow CS_2_{(l)}+2H_2S_{(g)}

The mole ratio of the reactants = 1 : 4

It means

0.5 moles of methane react with 2.0 moles of sulfur to give 0.5 moles of Carbon disulfide and 1.0 moles of hydrogen sulfide gas.

But available moles of S = 0.5 moles

Limiting reagent is the one which is present in small amount. Thus, S is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

4 moles of S produces 1 mole of CS_2

Thus,

0.5 moles of S produces \frac{1}{4}\times 0.5 mole of CS_2

<u>Mole of CS_2 = 0.125 mole</u>

Molar mass of CS_2 = 76.139 g/mol

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4 moles of S produces 2 moles of H_2S

Thus,

0.5 moles of S produces \frac{2}{4}\times 0.5 mole of H_2S

<u>Mole of H_2S = 0.25 mole</u>

Molar mass of H_2S = 34.1 g/mol

<u>Mass = Moles * Molar mass = 0.25 * 34.1 g = 8.525 g</u>

<u></u>

7 0
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7 0
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