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3 years ago

Question 11 (5 points)

Chemistry

1 answer:

LenKa [72]3 years ago

8 0

Answer:

False

Explanation:

On the left side of the equation (Li + O2), there is 1 Li atom and 2 O atoms.

but on thw right side of the equation (Li2O,) there is 2 Li atoms and 1 O atom

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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

Romashka [77]

Answer: The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

Explanation:

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

$W_{solvent}$ = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

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

Calculation for freezing point of solution:

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

Calculation for boiling point of solution:

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

8 0

3 years ago

The table shows the total number of electrons in Atom A and Atom B. Atom Number of Electrons A 10 B 18 Which statement is correc

HACTEHA [7]

Answer:

Both A and B will be unreactive!

Explanation:

10=2, 8

18=2, 8, 8

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What does the symbol "E3" represent?

Anna71 [15]

Answer:

your answer to your question is B

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An example of a solution that is formed endothermically is the one used in

Makovka662 [10]

Cold Packs

Explanation:

Endothermic reactions are reactions in which heat is absorbed from the surroundings and the container in which the reaction takes place is said to be very cold. In cold packs, ammonium nitrate is used, is dissolved in solution is said to be an example of endothermic reaction, in this the heat energy is absorbed from the surroundings. In this the solute increases the solubility with the rise in temperature, and also used in cold packs.

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Is a specimen a monocot or dicot

den301095 [7]

Monocots have only one seed leaf inside the seed coat. It is often only a thin leaf, because the endosperm to feed the new plant is not inside the seed leaf. Dicots have two seed leaves inside the seed coat. They are usually rounded and fat, because they contain the endosperm to feed the embryo plant.

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