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3 years ago

Calculate the volume in milliliters of a 5.55-g sample of a solid with a density of 3.03 g/mL

Chemistry

1 answer:

Marina86 [1]3 years ago

8 0

(4.99 g) / (3.65 g/mL) = 1.37 mL

The answer to your question is 1.37ml.

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Maple syrup, which comes from the sap of maple trees, contains water and natural sugars. It's a clear, brown liquid and the suga

WITCHER [35]

Maple syrup, which comes from the sap of maple trees, contains water and natural sugars. It's a clear, brown liquid and the sugars can’t be separated byfiltration. The category is insoluble.

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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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3 years ago

Photosynthesis uses all of the following except __ to make food ?

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What are the options

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3 years ago

8.310x10^2 – 7.210x10^1 [?]x10^[?]

alisha [4.7K]

Answer:

$7.589\times 10^{2}$

Explanation:

The expression can be solved mathematically as follows:

1) $8.310\times 10^{2}-7.210\times 10^{1}$ Given

2) $8.310\times 10^{1+1} - 7.210\times 10^{1}$ Definition of sum

3) $(8.310\times 10^{1})\times 10^{1}-7.210\times 10^{1}$ $a^{m+n} = a^{m}\cdot a^{n}$ /Associative property

4) $(8.310\times 10^{1}-7.210)\times 10^{1}$ Distributive property

5) $(83.100-7.210)\times 10^{1}$ Multiplication

6) $75.89\times 10^{1}$ Subtraction.

7) $(7.589\times 10^{1})\times 10^{1}$ Multiplication/Associative property

8) $7.589\times (10^{1}\times 10^{1})$ Associative property

9) $7.589\times 10^{2}$ $a^{m+n} = a^{m}\cdot a^{n}$ /Result

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3 years ago

Why is argon and chlorine both gases at room temperature?

vredina [299]

Argon and chlorine are both gases at room temperature because they are non-metals.

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