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3 years ago

4 pts

1 answer:

7 0

First we need to find the number of moles that 43.9g of gallium metal is. We can do this by finding the molar weight of gallium and cross-multiplying to cancel out units:

$\frac{1mole}{69.723g} *\frac{43.9g}{1} =0.63MOLga$

So we are dealing with 0.63 moles of gallium metal.

We can take from the balanced equation that 4 moles of gallium metal will react completely with 3 moles of oxygen gas. We can take this ratio and make a proportion to find the amount of oxygen gas, in moles, that will react completely with 0.63 moles of gallium metal:

$\frac{4molesGa}{3moles02} =\frac{0.63molesga}{xmoleso2}$

Cross multiply and solve for x:

4x=1.89

x=47 molesO₂

So now we know that 0.47 moles of oxygen gas will react with 43.9g of gallium metal.

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Ice has a density of 0.92g/cm3. It will float in water.

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The density of ice is about 90 percent that of water, but that could range because ice can contain air, too. meaning that about 10 percent of an ice cube or iceberg will be above the water line. The density of water is maximum at four∘C, and the density of the ice is much less than the water due to its susceptible intermolecular pressure of attraction. as the density of water is more, it's miles heavier than ice. therefore ice floats on the floor of the water. Ice continually floats due to the fact it's far less dense than everyday water. because frozen water molecules shape a crystal.

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Any substance that when dissolved in water will give off h ions is referred to as a(n) ________.

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3 years ago

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

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