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loris [4]
3 years ago
15

A block of metal has a mass of 18.361kg and the following dimensions: 22.35 cm x 7.34 cm x 22.05 mm. What is the density of the

metal in g/cm^3
Chemistry
1 answer:
professor190 [17]3 years ago
3 0

Answer:

50.76 g/cm³

Explanation:

We'll begin by converting 18.361 Kg to g. This can be obtained as follow:

1 Kg = 1000 g

Therefore,

18.361 Kg = 18.361 Kg × 1000 g / 1 Kg

18.361 Kg = 18361 g

Next, we shall express the dimension in the same unit of measurement.

Dimension = 22.35 cm x 7.34 cm x 22.05 mm

Thus, we shall convert 22.05 mm to cm. This can be obtained as follow:

10 mm = 1 cm

Therefore,

22.05 mm = 22.05 mm × 1 cm / 10 mm

22.05 mm = 2.205 cm

Thus, the dimension is 22.35 cm x 7.34 cm x 2.205 cm.

Next, we shall determine the volume of the metal. This can be obtained as follow:

Dimension = 22.35 cm x 7.34 cm x 2.205 cm.

Volume = 361.728045 cm³

Finally, we shall determine the density of the metal. This can be obtained as follow:

Mass of metal = 18361 g

Volume of metal = 361.728045 cm³

Density of metal =?

Density = mass / volume

Density of metal = 18361 / 361.728045

Density of metal = 50.76 g/cm³

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Answer:

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8 0
3 years ago
An aqueous CsCl solution is 8.00 wt% CsCl and has a density of 1.0643 g/mL at 20°C. What is the boiling point of this solution?
umka2103 [35]

<u>Answer:</u> The boiling point of solution is 100.53

<u>Explanation:</u>

We are given:

8.00 wt % of CsCl

This means that 8.00 grams of CsCl is present in 100 grams of solution

Mass of solvent = (100 - 8) g = 92 grams

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 solution = 100°C

i = Vant hoff factor = 2 (For CsCl)

K_b = molal boiling point elevation constant = 0.51°C/m

m_{solute} = Given mass of solute (CsCl) = 8.00 g

M_{solute} = Molar mass of solute (CsCl) = 168.4  g/mol

W_{solvent} = Mass of solvent (water) = 92 g

Putting values in above equation, we get:

\text{Boiling point of solution}-100=2\times 0.51^oC/m\times \frac{8.00\times 1000}{168.4g/mol\times 92}\\\\\text{Boiling point of solution}=100.53^oC

Hence, the boiling point of solution is 100.53

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