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pychu [463]
3 years ago
15

In an experiment, hydrochloric acid reacted with different volumes of sodium thiosulfate in water. A yellow precipitate was form

ed during the reaction. A cross drawn at the base of each flask became gradually invisible due the formation of this yellow precipitate. The time taken for the cross to become invisible was recorded. A partial record of the experiment is shown.
Experimental Record Flask Volume of

HCl Volume of

Sodium Thiosulfate Volume of

Water Time

1 10 mL 10 mL 40 mL 14 seconds

2 10 mL 20 mL 30 mL

3 10 mL 30 mL 20 mL

4 10 mL 40 mL 10 mL


Based on your knowledge of factors that affect the rates of chemical reactions, predict the trend in the last column of the experimental record. Use complete sentences to explain the trend you predicted. You do not have to determine exact values for time; just describe the trend you would expect (increase or decrease) and why it occurs
Chemistry
1 answer:
morpeh [17]3 years ago
5 0

Answer:

The time taken for the cross to become invisible decreases.

Explanation:

We know that one of the factors affecting the rate of reaction is the concentration of reactants. From the collision theory, we know that the higher the concentration of reactants, the greater the possibility of effective collision between reactants leading ultimately to an increase in the rate of reaction. Increase in the rate of reaction implies that the reaction takes a shorter time to reach completion.

In the case of the reaction shown in the question, the point when the reaction is completed is observed by the time take for the cross mark to become invisible. If we look at the given data closely, we will notice that the volume of acid was held constant, the volume of thiosulphate was increased gradually while the volume of water was decreased accordingly. This implies that the concentration of the reactants was increased. Decreasing the volume of water increases reactant concentration.

As explained above, increase in reactant concentration increases the rate of reaction. Hence, the rate of reaction of the acid and thiosulphate increases as reactant concentration increases and the cross mark becomes invisible faster. This implies that in the last column for time taken for the cross to become invisible, the values of time decreases steadily as concentration of reactants increases.

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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
3 years ago
A sample of oxalic acid (a diprotic acid of the formula H2C2O4) is dissolved in enough water to make 1.00 L of solution. A 100.0
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<u>Answer:</u> The mass of original oxalic acid sample is 6.75 grams

<u>Explanation:</u>

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n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2C_2O_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=?M\\V_1=100.0mL\\n_2=1\\M_2=0.750M\\V_2=20.0mL

Putting values in above equation, we get:

2\times M_1\times 100.0=1\times 0.750\times 20.0\\\\M_1=\frac{1\times 0.750\times 20.0}{2\times 100.0}=0.075M

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\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

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Putting values in above equation, we get:

0.075M=\frac{\text{Mass of oxalic acid}}{90g/mol\times 1L}\\\\\text{Mass of oxalic acid}=(0.075\times 90\times 1)=6.75g

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