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2 years ago

There are ??? molecules in one mole regardless of what the substance is.

1 answer:

8 0

Answer: The mole allows people to calculate the number of middle schoolers entities (usually atoms or molecules. Isabella's number is an absolute number: there are 6.022 × 1023 middle schoolers entities is in 1 mole. This can also be written as 6.022 × 1023 mol-1.

Explanation: I hope that helped !!

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Margaret [11]

Answer:

Recycling and reuse of materials

Explanation:

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2 years ago

Do the molecules in a desk have energy?

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2 years ago

A solution made by mixing 20.0 g of a non-volatile compound with 125 mL of water at 25°C has a vapor pressure of 22.67 torr. Wha

Ainat [17]

We have that the molecular weight (3sf) of the compound (g/mol)

$m=44.15g/mol$

From the question we are told

A solution made by mixing 20.0 g of a non-volatile compound with 125 mL of water at 25°C has a vapor pressure of 22.67 torr. What is the molecular weight (3sf) of the compound (g/mol).

Generally the equation for the Rouault's law is mathematically given as

P=P_0 N

$22.67=23.8*\frac{\frac{12.5}{18}}{\frac{125}{18}+\frac{15}{m}}\\\\\6.95+\frac{15}{m}=7.29\\\\\frac{15}{m}=7.29-6.95\\\\m=\frac{15}{0.34}\\\\m=44.11g/mol$

Therefore

The molecular weight (3sf) of the compound (g/mol)

$m=44.15g/mol$

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1 year ago

a friend is trying to describe a material to you. which of the following is not a description of a physical property of the subs

jonny [76]

C. when combined with water, it fizzes.

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2 years ago

Read 2 more answers

When 74.8g of alanine C3H7NO2 are dissolved in 1450.g of a certain mystery liquid X, the freezing point of the solution is 8.30°

dlinn [17]

Answer: The mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=8.30^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte)

$K_f$ = freezing point constant =

m= molality = $\frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}=\frac{74.8g\times 1000}{1450g\times 89.09g/mol}=0.579$

$8.30^0C=1\times K_f\times 0.579$

$K_f=14.3^0C/m$

Let Mass of solute (KBr) = x g

$8.3^0C=1.72\times 14.3\times \frac{xg\times 1000}{119g/mol\times 1450g}$

$x=58.2g$

Thus the mass of potassium bromide that must be dissolved in the same mass of X to produce the same depression in freezing point is 58.2 grams

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1 year ago