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shusha [124]
2 years ago
13

Which of the following solutions contains the largest number of moles of dissolved particles? a.) 25 ml of sodium chloride b.) 2

5 ml of aluminum hydroxide c.) 25 ml of acetic acid d.) 25 ml of nitric acid e.) 25 ml of sugar
Chemistry
2 answers:
kirza4 [7]2 years ago
5 0

Answer:

b.) 25 ml of aluminum hydroxide

Explanation:

For this question, we have to assume that we have the <u>same concentration</u> for all the solutions, for example, <u>1 M</u>. Additionally, we have to take into account the <u>ionization reaction</u> for each species:

a) NaCl~->~Na^+~+~Cl^- <u>we have to ions</u>

b) Al(OH)_3~->~Al^+^3~+~3OH^- <u>we have fourth ions</u>

c) CH_3COOH~->~CH_3COO^-~+~H^+ <u>we have two ions</u>

d) C_6H_1_2O_6~->~C_6H_1_2O_6  <u>we have one ion</u>

If we have the same volume and the same concentration the variables that will help us to answer the question would be the n<u>umber of ions.</u> If we have <u>more ions we will have more particles dissolved</u>. Therefore the answer would be b) (<u>due to the fourth ions</u>).

I hope it helps

katen-ka-za [31]2 years ago
4 0

Answer:

b.) 25 ml of aluminum hydroxide.

Explanation:

Hello,

In this case, bu knowing the formula of each substance, we can predict the number of particles are going to be dissolved in the given solutions, thus we have:

Sodium chloride, NaCl: when it is placed in water, two ions, sodium and chloride are dissolved.

Aluminium hydroxide, Al(OH)3: when it is placed in water, four ions are dissolved, one aluminium (III) and three hydroxyl.

Acetic acid CH3COOH: when it is placed in water, and equilibrium dissolution is present, therefore, not all the acetic acid particles will be dissolved as ions, anyway, two ionic species are formed, acetate anion and hydronium.

Sugar C12H22O11: when it is placed in water, it is not properly dissociated but it is said to form hydrogen bonds in order to get dissolved in water, therefore, just one dissolved particle is formed.

In such a way, the substance having the larges number of moles of dissolved particles is aluminium hydroxide having four, by considering the same concentration for all the solutions, say 1 M or that so.

Regards.

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0.53g of acetanilide was subjected to kjeldahl determination and the ammonia produced was collected in 50cm3 of 0.50M of h2so4.o
Lady bird [3.3K]

Answer:

10.57% of N in acetanilide

Explanation:

All nitrogen in the sample is converted in NH₃ in the Kjeldahl determination. The NH₃ reacts with H₂SO₄ as follows:

2NH₃ + H₂SO₄ → 2NH₄⁺ + SO₄²⁻

The acid in excess in titrated with Na₂CO₃ as follows:

Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂

To solve this question we must find the moles of sodium carbonate = Moles of H₂SO₄ in excess. The added moles - Moles in excess = Moles of sulfuric acid that reacts:

<em>Moles Na₂CO₃ anf Moles H₂SO₄ in excess:</em>

0.025L * (0.05mol / L) = 1.25x10⁻³ moles Na₂CO₃ / 0.01360L =

0.09191M * 0.250L = 0.0230 moles H₂SO₄ in excess.

<em>Moles H₂SO₄ added:</em>

0.050L * (0.50mol / L) = 0.0250 moles H₂SO₄ added

<em>Moles that react:</em>

0.0250 moles - 0.0230 moles = 0.0020 moles H₂SO₄

<em>Moles of NH₃ = Moles N:</em>

0.0020 moles H₂SO₄ * (2mol NH₃ / 1mol H₂SO₄) = 0.0040 moles NH₃ = Moles N

<em>mass N and mass percent:</em>

0.0040 moles N * (14g / mol) = 0.056gN / 0.53g * 100 =

<h3>10.57% of N in acetanilide</h3>
7 0
2 years ago
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