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3 years ago

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:

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.

So, the right choice is: D.

Rom4ik [11]3 years ago

8 0

D on e2020

extra words to get to 20 letters

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According to boyle p1v1=p2v2
So 3*4=6*x so x=2

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What is the oxidation state of CI in CIO- -1 +2 +1 0

puteri [66]

Answer:

Explanation:

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Draw a mechanism for the reaction of methylamine with 2-methylpropanoic acid. Draw any necessary curved arrows. Show the product

Nitella [24]

Answer:

See figure 1

Explanation:

On this case we have a base (methylamine) and an acid (2-methyl propanoic acid). When we have an acid and a base an acid-base reaction will take place, on this specific case we will produce an ammonium carboxylate salt.

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Due to the nature of the compounds (base and acid), the nucleophile (methylamine) doesn't have the ability to attack the carbon of the carbonyl group due to his basicity. The methylamine will react with the acid-producing a positive charge on the nitrogen and with this charge, the methylamine loses all his nucleophilicity.

I hope it helps!

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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, Moles of ammonium chloride formed = 0.5 moles

Molar mass of ammonium chloride = 53.491 g/mol

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(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 $CS_2$

Mole of $CS_2$ = 0.125 mole

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 $H_2S$

Mole of $H_2S$ = 0.25 mole

Molar mass of $H_2S$ = 34.1 g/mol

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