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3 years ago

If 3.50 moles of sodium chloride is added to your food, how many grams is added?

Chemistry

1 answer:

Ksenya-84 [330]3 years ago

7 0

Answer:

204.8g

Explanation:

The number of moles of a substance is related to its mass and molecular mass as follows:

mole (n) = mass (m) ÷ molar mass (MM)

According to this question, 3.50 moles of sodium chloride (NaCl) is added to a food.

Molar mass of NaCl = 23 + 35.5

= 58.5g/mol

Using mole = mass/molar mass

Mass = molar mass × mole

Mass = 58.5g/mol × 3.5mol

Mass = 204.75

Mass = 204.8grams.

Therefore, 204.8grams of NaCl or common salt was added to the food.

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3 years ago

Read 2 more answers

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

sammy [17]

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

What makes the periodic table arrangement so unique and useful?

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Elements are ordered from left to right, increasing atomic number as they go. The periodic table allows us to find out and predict things about elements that haven’t been discovered yet!

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3 years ago

If the concentrated form of blood expander is 9.0 mol/L and one of your companions, a nurse, tells you that you need to have a f

I am Lyosha [343]

Answer:

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Explanation:

Use the dilution equation M1V1 = M2V2

M1 = 9.0 M

V1 = This is what we're looking for.

M2 = 0.145 M

V2 = 2 L

Solve for V1 --> V1 = M2V2/M1

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Answer:

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