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2 years ago

Rank the following steps involved in preparing a dilute solution from a stock solution in order from first to last.

Chemistry

1 answer:

Serggg [28]2 years ago

5 0

Answer: C - B - A - D

Explanation:

-Obtain a flask that is thoroughly cleaned

-Add the desired amount of stock solution

-Add deionized water until the volume reaches the mark on the flask.

- Cap the flask and invert several times to thoroughly mix the solution.

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The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 2.60 g of naphthalene (C10H8) in 675 g of be

Mrac [35]

Answer: The freezing point of solution is 5.35°C

Explanation:

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

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

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 4.90°C/m

$m_{solute}$ = Given mass of solute (naphthalene) = 2.60 g

$M_{solute}$ = Molar mass of solute (naphthalene) = 128.2 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 675 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 4.90^oC/m\times \frac{2.60\times 1000}{128.2g/mol\times 675}\\\\\text{Freezing point of solution}=5.35^oC$

Hence, the freezing point of solution is 5.35°C

3 0

2 years ago

The solubility of carbon dioxide in water is 0.161 g CO2in 100 mL of water at 20ºC and 1.00 atmCO2. A soft drink is carbonated w

Sedbober [7]

Answer: The solubility of carbon dioxide at 5.50 atm is $0.886g/100mL$

Explanation:

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{A}=K_H\times p_{A}$

Or,

$\frac{C_{1}}{C_{2}}=\frac{p_{1}}{p_2}$

where,

$C_1\text{ and }p_1$ are the initial concentration and partial pressure of carbon dioxide

$C_2\text{ and }p_2$ are the final concentration and partial pressure of carbon dioxide

We are given:

$C_1=0.161g/100mL\\p_1=1.00atm\\C_2=?\\p_2=5.50atm$

Putting values in above equation, we get:

$\frac{0.161g/100mL}{C_2}=\frac{1.00atm}{5.50atm}\\\\C_2=\frac{0.161g/100mL\times 5.50atm}{1.00atm}=0.886g/100mL$

Hence, the solubility of carbon dioxide at 5.50 atm is $0.886g/100mL$

4 0

3 years ago

How many lone pairs are around the central atom in the Lewis formula for the

barxatty [35]

Answer:

The answer to your question is None, sulfur share of its electrons

Explanation:

Just remember:

Sulfur, S, has 6 electrons in its outermost shell

Hydrogen, H, has 1 electron in its outermost shell

Oxygen, O, has 6 electrons in its outermost shell

See the picture below

The electrons of sulfur are in blue

The electrons of oxygen are in red

The electron in hydrogen is in yellow

Sulfur is the central atom and it shares all its electrons with the oxygen.

8 0

3 years ago

How is a pure substance different from a mixture?

Sauron [17]

The answer is A.
Pure substance is either an element or a compound, which elements in compund is chemically combined together. They cannot be separated by physical methods such as filtration or evaporation. Compounds can only be separated by chemical methods, which include using electricity (electrolysis) or applying heat.

7 0

2 years ago

Read 2 more answers

What would be the properties of the 5th state of matter? (bosse-instein condensate)

kolbaska11 [484]

Answer:

a state of matter created when particles, called bosons, are cooled to near absolute zero (-273.15 degrees Celsius, or -460 degrees Fahrenheit).

Explanation:

hope this helps

pls mark brainliest

8 0

2 years ago