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3 years ago

An object has a mass of 48g and it has a height of 5cm, a width of 2cm and a length of 5cm.

Chemistry

1 answer:

Ainat [17]3 years ago

3 0

Answer:

0.96g/cm³

Explanation:

Given parameters:

Mass of the object = 48g

Height of the object = 5cm

Width of the object = 2cm

Length of the object = 5cm

Unknown:

Density = ?

Solution:

Density of a body is the mass per unit volume of the body.

Mass is the amount of matter in a substance

Volume is the space such a body occupies

To find the volume;

Volume = l x w x h

Insert the parameters and solve;

Volume = 5cm x 2cm x 5cm = 50cm³

Density = $\frac{mass}{volume}$

Density = $\frac{48 }{50}$ = 0.96g/cm³

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A solid is 5 cm tall 3 cm wide and 2 cm thick it has a mass of 129 g what is the density

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A solution is prepared by dissolving 27.0 g of urea [(NH2)2CO], in 150.0 g of water. Calculate the boiling point of the solution

andrew11 [14]

Answer: The boiling point of solution is 101.56°C

Explanation:

Elevation in boiling point is defined as the difference in the boiling point of solution and boiling point of pure solution.

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

Or,

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

where,

Boiling point of pure water = 100°C

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

$K_b$ = molal boiling point elevation constant = 0.52°C/m.g

$m_{solute}$ = Given mass of solute (urea) = 27.0 g

$M_{solute}$ = Molar mass of solute (urea) = 60 g/mol

$W_{solvent}$ = Mass of solvent (water) = 150.0 g

Putting values in above equation, we get:

$\text{Boiling point of solution}-100=1\times 0.52^oC/m\times \frac{27\times 1000}{60\times 150}\\\\\text{Boiling point of solution}=101.56^oC$

Hence, the boiling point of solution is 101.56°C

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4 years ago

What is the word for mixture of solids and water, plz an accurate answer, otherwise report.

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Water is H2O but I'm not sure about solids.

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3 years ago

A 150.0 mL sample of an aqueous solution at 25°C contains 15.2 mg of an unknown nonelectrolyte compound. If the solution has an

inysia [295]

Answer: The molar mass of the unknown compound is 223.2 g/mol

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=iMRT$

where,

$\pi$ = osmotic pressure of the solution = 8.44 torr

i = Van't hoff factor = 1 (for non-electrolytes)

M = molarity of solute = ?

R = Gas constant = $62.3637\text{ L torr }mol^{-1}K^{-1}$

T = temperature of the solution = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$8.44torr=1\times M\times 62.3637\text{ L. torr }mol^{-1}K^{-1}\times 298K\\\\M=\frac{8.44}{1\times 62.3637\times 298}=4.54\times 10^{-4}M$

To calculate the molecular mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Molarity of solution = $4.54\times 10^{-4}M$

Given mass of unknown compound = 15.2 mg = 0.0152 g (Conversion factor: 1 g = 1000 mg)

Volume of solution = 150.0 mL

Putting values in above equation, we get:

$4.54\times 10^{-4}M=\frac{0.0152\times 1000}{\text{Molar mass of unknown compound}\times 150.0}\\\\\text{Molar mass of unknown compound}=\frac{0.0152\times 1000}{150.0\times 4.54\times 10^{-4}}=223.2g/mol$

Hence, the molar mass of the unknown compound is 223.2 g/mol

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3 years ago

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