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andreyandreev [35.5K]
3 years ago
11

In which situation are the individual molecules moving the fastest? A) in a bowl of hot soup B) in a glass of iced tea C) in a g

lass of tap water D) in a cup of boiling water\
Chemistry
2 answers:
Airida [17]3 years ago
5 0
<h2>Answer : Option D) In a cup of boiling water</h2><h3>Explanation :</h3>

The individual molecules of water while boiling will move fastest amongst the given choices. As the boiling of water will cause the liquid molecules to increase the intermolecular forces of attraction between molecules and get escaped into vapors. Here, there is a transformation of state that is about to occur from liquid to gaseous. Compared to other examples, where there are no change of state occurring.

tresset_1 [31]3 years ago
5 0

D) in a cup of boiling water

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A scientist prepared an aqueous solution of a 0.45 M weak acid. The pH of the solution was 2.72. What is the percentage ionizati
aleksandr82 [10.1K]

Answer:

0.42%

Explanation:

<em>∵ pH = - log[H⁺].</em>

2.72 = - log[H⁺]

∴ [H⁺] = 1.905 x 10⁻³.

<em>∵ [H⁺] = √Ka.C</em>

∴ [H⁺]² = Ka.C

∴ ka = [H⁺]²/C = (1.905 x 10⁻³)²/(0.45) = 8.068 x 10⁻⁶.

<em>∵ Ka = α²C.</em>

Where, α is the degree of dissociation.

<em>∴ α = √(Ka/C) </em>= √(8.065 x 10⁻⁶/0.45) = <em>4.234 x 10⁻³.</em>

<em>∴ percentage ionization of the acid = α x 100</em> = (4.233 x 10⁻³)(100) = <em>0.4233% ≅ 0.42%.</em>

4 0
3 years ago
Read 2 more answers
Describe what happens in a condensation reactions
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A condensation reaction is described to be a reaction wherein two molecules form an even larger product and consequently produces a smaller molecule as a by-product. For example, when two amino acids are combined, a dipeptide bond is formed. As a result, 1 molecule of water is produced as a by-product.
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When aqueous solutions of manganese(II) iodide and sodium phosphate are combined, solid manganese(II) phosphate and a solution o
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Answer:

The net ionic equation for the given reaction :

3Mn^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)

Explanation:

3MnI_2(aq)+2Na_3PO_4_2(aq)\rightarrow Mn_3(PO_4)_2(s)+6NaI(aq)...[1]

MnI_2(aq)\rightarrow Mn^{2+}(aq)+2I^-(aq)..[2]

Na_3PO_4(aq)\rightarrow 3Na^{+}(aq)+PO_4^{3-}(aq)...[3]

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Replacing MnI_2(aq) , NaI and Na_3PO_4(aq) in [1] by usig [2] [3] and [4]

3Mn^{2+}(aq)+6I^-(aq)+6Na^{+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)+6Na^+(aq)+6I^-(aq)

Removing the common ions present ion both the sides, we get the net ionic equation for the given reaction [1]:

3Mn^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)

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Explanation:

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Explanation:

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