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Anarel [89]
3 years ago
15

Element X reacts with hydrogen gas at 200°C to form compound Y. When Y is heated to a higher temperature, it decomposes to the e

lement X and hydrogen gas in the ratio of 559 mL of H2 (measured at standard temperature and pressure) for 1.00 g of X reacted. X also combines with chlorine to form a compound Z, which contains 63.89 percent by mass of chlorine. Deduce the identity of X. The symbol of element X is
Chemistry
1 answer:
vfiekz [6]3 years ago
5 0

Answer:

X is calcium with the symbol Ca

Explanation:

From the first statement:

X reacted with H2 to produce Z i.e

X + H2 —> Y

Y is heated to form X and H2 i.e

Y —> X + H2

Let us obtain moles of H2 produced. This is illustrated below:

Volume of H2 produced = 559 mL = 0.559 L.

I mole of H2 occupy 22.4L at stp

Therefore, b mol of H2 will occupy 0.559 L at stp i.e

b mol of H2 = 0.559/22.4

b mol of H2 = 0.025 mole

Next 0.025 mole of H2 to gram..

Molar Mass of H2 = 2x1 = 2g/mol

Mole of H2 = 0.025 mole

Mass = number of mole x molar Mass

Mass of H2 = 0.025 x 2

Mass of H2 = 0.05g

From the question given, we were told that 1g of X reacted. This means that 1g of X reacted with 0.05g of H2. Now let us determine the mass of X that will react with 1 mol ( i.e 2g) of H2. This is illustrated below:

From the reaction,

It was discovered that 1g of X reacted with 0.05g of H2.

Therefore, P g of X will react with 2g of H2 i.e

P g of X = 2/0.05 = 40g/mol

The molar mass of X is 40g/mol.

Now let us consider the second statement to see if we'll obtained the same result as 40g/mol of X

From the second statement:

X also combines with chlorine to form a compound Z, which contains 63.89 percent by mass of chlorine i.e

X + Cl2 —> Z

Molar Mass of Z (X + 2Cl) = X + (35.5 x 2) = X + 71.

Z contains 63.89% by mass of Cl2.

We can obtain the molar mass of X as follow:

Percentage by mass of Cl2 in the compound Z is given by:

Mass of Cl2/Molar Mass x100

63.89/100 = 71/(X + 71)

Cross multiply to express in linear form

63.89(X + 71) = 100 x 71

Clear the bracket

63.89X + 4536.19 = 7100

Collect like terms

63.89X = 7100 - 4536.19

63.89X = 2563.81

Divide both side by 63.89

X = 2563.81/63.89

X = 40g/mol

Now we can see that in both experiments, the molar mass of X is 40g/mol.

Comparing the value of X i.e 40g/mol with that from the periodic table, X is calcium with the symbol Ca

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A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and
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<u>Answer:</u> The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}

Where,

m_{solute} = Given mass of solute (S_8) = 0.800 g

M_{solute} = Molar mass of solute (S-8) = 256.52 g/mol

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Putting values in above equation, we get:

\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m

  • <u>Calculation for freezing point of solution:</u>

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}

To calculate the depression in freezing point, we use the equation:

\Delta T_f=iK_fm

or,

\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

K_f = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC

Hence, the freezing point of solution is 16.5°C

  • <u>Calculation for boiling point of solution:</u>

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

or,

\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

K_f = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC

Hence, the boiling point of solution is 118.2°C

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