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2 years ago

A compound has a molar mass of 44.01 g/mol. What is its identity?

1 answer:

6 0

This problem is very vague since no other details are given. However for the sake of calculation let us assume that the compound is an ideal gas at STP.

So that the molar volume is: 1 mole = 22.4 L

density = (44.01 g/mol) * (1 mol / 22.4 L) = 1.96 g/L = 1.96 kg/m^3

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Soils that allow water to pass through them faster are more

Kay [80]

The answer would be permeable because permeable means let's liquid through easily.

7 0

2 years ago

Read 2 more answers

You are given a food containing 6 g protein per serving (30g), how much of the food you needto add to 100 ml solution to make a

Bas_tet [7]

Hey there!

500 mg of protein is present in 100 mL of solvent as per the concentration 0.5 mg/mL or 500 g/mL ,

So, 250 mg (0.25 g) of serving food need to be added to 100 mL solvent in order to prepare 50 mg of protein/100 mL solution.

Dilution factor = initial amount of protein / final amount of protein

= 6 g / 0.05 g = 120

Hope this helps!

4 0

3 years ago

Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depres

yulyashka [42]

Answer: The freezing point of solution is -0.454°C

Explanation:

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

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 = 0°C

i = Vant hoff factor = 2

$K_f$ = molal freezing point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute (KCl) = 5.0 g

$M_{solute}$ = Molar mass of solute (KCl) = 74.55 g/mol

$W_{solvent}$ = Mass of solvent (water) = 550.0 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{74.55g/mol\times 550}\\\\\text{Freezing point of solution}=-0.454^oC$

Hence, the freezing point of solution is -0.454°C

3 0

2 years ago

If there are 7 moles of copper dissolved in 15 moles of zinc to make 22 moles of a brass solution. What is the mole fraction of

iragen [17]

Mole fraction of copper is 7/22

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3 years ago

Which property determines an atom's ability to attract electrons shared in a chemical bond?

ASHA 777 [7]

It is electronegativity.

4 0

3 years ago