Chemistry College

2. 4.6gof X is burnt completelyto produce 6.2g of X oxide (X,O). M (0) = 16 gmol ¹. Calculate the amount of oxygen that reacted in this experiment. [2 MARKS]
[ii] calculate the mass of 1 mole of x.[2mark]
[iii] predict and give a reason explaining the reaction of x2o in water.[1mark]​

Answers

Answer 1

As per the given data, 1.6 grams of oxygen reacted in this experiment.

To calculate the amount of oxygen that reacted in the experiment, we need to determine the difference in the mass of X oxide (X,O) formed and the mass of X initially used.

Given:

Mass of X = 4.6 g

Mass of X oxide (X,O) = 6.2 g

To find the amount of oxygen that reacted:

Mass of oxygen = Mass of X oxide - Mass of X

= 6.2 g - 4.6 g

= 1.6 g

Therefore, 1.6 grams of oxygen reacted in this experiment.

Calculate the mass of 1 mole of X:

Given that the mass of X is 4.6 g, we can calculate the molar mass of X by dividing the mass by the number of moles:

Molar mass of X = Mass of X / Number of moles of X

Molar mass of X = 4.6 g / 0.1 mol

Molar mass of X = 46 g/mol

Therefore, the mass of 1 mole of X is 46 grams.

Thus, the answer is 46 grams.

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Related Questions

The pH of an acidic solution is 2.83. What is [H*]?

Answers

Answer:

0.001464 M, or 1.464 × 10^(-3) M.

Explanation:

[H+] = 10^(-pH)

In this case, the pH of the acidic solution is 2.83. Plugging this value into the equation, we get:

[H+] = 10^(-2.83)

Using a calculator, we can find that 10^(-2.83) is approximately 0.001464.

Therefore, the concentration of hydrogen ions in the acidic solution is approximately 0.001464 M, or 1.464 × 10^(-3) M.

The following reactions (note that the arrows are pointing only one direction) can be used to prepare an activity series for the halogens:
Br2(aq)+2NaI(aq)⟶2NaBr(aq)+I2(aq)

Cl2(aq)+2NaBr(aq)⟶2NaCl(aq)+Br2(aq)
A) Predict whether a reaction will occur when elemental chlorine and potassium bromide are mixed.
Express your answer as a chemical equation.

B)Predict whether a reaction will occur when elemental iodine and lithium chloride are mixed.
Express your answer as a chemical equation.

Answers

A) The chemical equation for the reaction is [tex]Cl_2(aq) + 2KBr(aq)[/tex] ⟶ [tex]2KCl(aq) + Br_2(aq)[/tex]

B)No reaction occurs when elemental iodine is mixed with lithium chloride. [tex]I_2(aq) + 2LiCl(aq)[/tex]⟶ No Reaction

A) To predict whether a reaction will occur when elemental chlorine ([tex]Cl_2[/tex]) and potassium bromide (KBr) are mixed, we can refer to the activity series for the halogens. According to the activity series, chlorine is more reactive than bromine. Therefore, chlorine can displace bromine from its compounds.

The chemical equation for the reaction between chlorine and potassium bromide can be written as:

[tex]Cl_2(aq) + 2KBr(aq)[/tex] ⟶ [tex]2KCl(aq) + Br_2(aq)[/tex]

In this reaction, chlorine displaces bromine from potassium bromide, resulting in the formation of potassium chloride and elemental bromine.

B) Similarly, to predict whether a reaction will occur when elemental iodine ([tex]l_2[/tex]) and lithium chloride (LiCl) are mixed, we can refer to the activity series. In the halogen activity series, iodine is less reactive than chlorine and bromine. Therefore, it is less likely for iodine to displace chlorine or bromine from their compounds.

The chemical equation for the reaction between iodine and lithium chloride can be written as:

[tex]I_2(aq) + 2LiCl(aq)[/tex]⟶ No Reaction

No reaction occurs because iodine is less reactive than chlorine, and lithium chloride does not react with iodine under these conditions.

Therefore, when elemental chlorine is mixed with potassium bromide, a reaction occurs and chlorine displaces bromine. On the other hand, no reaction occurs when elemental iodine is mixed with lithium chloride.

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The OH concentration in an aqueous solution at 25 °C is 3.3 x 10³.
What is [H*]?

Answers

The concentration of hydroxide ions ([OH-]) and the concentration of hydronium ions ([H+]) are related in an aqueous solution by the equation [H+][OH-] = 1.0 x 10^-14 at 25 °C .The concentration of hydronium ions ([H+]) in the aqueous solution at 25 °C is approximately 3.03 x 10^-18.

Given that [OH-] is 3.3 x 10^3, we can substitute this value into the equation as follows:

[H+][3.3 x 10^3] = 1.0 x 10^-14

Dividing both sides of the equation by 3.3 x 10^3, we get:

[H+] = (1.0 x 10^-14) / (3.3 x 10^3)

Simplifying the expression, we have:

[H+] ≈ 3.03 x 10^-18

In summary, at 25 °C, an aqueous solution with an OH- concentration of 3.3 x 10^3 has a hydronium ion concentration of approximately 3.03 x 10^-18. The hydronium ion concentration is determined by the equilibrium constant for water dissociation and is inversely proportional to the hydroxide ion concentration. The two concentrations are related through the equation [H+][OH-] = 1.0 x 10^-14.

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