Question

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Answer 1

Answers in solutions.

Step-by-step explanation:

Question 6:

The density of gold is 19.3 g/cm³

The density of silver is 10.5 g/cm³

• The density of the substance in Crown A;

Density = mass ÷ volume =
`(1930)/(100)` = 19.3 g/cm³

Since the density of gold, given, is 19.3 g/cm³ and the density of the substance in Crown A has a density of 19.3 g/cm³ , then that substance must be gold.

• The density of the substance in Crown B;

Density = mass ÷ volume = 1930 ÷ 184 = 10.48913043 g/cm³ ≈ 10.5 g/cm³ (answer rounded up to one decimal place)

Since the density of the substance in Crown B is approximately equal to 10.5 g/cm³ , then that substance is Silver.

• The density of substance in Crown C;

Density = mass ÷ volume = 1930g ÷ 150cm³ = 12.86666667 ≈ 12.9 cm³ (answer rounded up to one decimal place)

The density of the mixture:

For 2 cm³ of the mixture, its mass equal 19.3 g + 10.5 g = 29.8 g

∴ for 1 cm³ of the mixture, its mass equal to
`(29.8)/(2)` = 14.9 g

Hence the density of the mixture = 14.9 g/cm³ and is not equal to the density of the substance in Crown C.

* Crown C is not made up of a mixture of gold and silver.

Question 7:

• An empty masuring cylinder has a mass of 500 g.

• Water is poured into measuring cylinder until the liquid level is at the 100 cm³ mark.

• The total mass is now 850 g

The mass of water that occupied the 100 cm³ space of the container = total mass - mass of the empty container = 850 g - 500 g = 350 g

Density of the liquid (water) poured into the container = mass ÷ volume = 350 g ÷ 100 cm³ = 3.5g/cm³

Question 8:

A tank filled with water has a volume of 0.02 m³

(a) 1 liter = 0.001 m³

How many liters? = 0.02 m³ ?

Cross multiplying gives:

`(0.02 * 1)/(0.001)` = 20 liters

(b) 1 m³ = 1,000,000 cm³

0.02 m³ = how many cm³ ?

Cross-multiplying gives;

`(0.02 * 1,000,000)/(1)` = 20,000 cm³

(c) 1 cm³ = 1 ml

∴ 0.02 m³ of the water = 20,000 cm³ = 20,000 ml

Question 9:

Caliper (a) measurement = 3.2 cm

Caliper (b) measurement = 3 cm

Question 10:

• A stone is gently and completely immersed in a liquid of density 1.0 g/cm³
• in a displacement can
• The mass of liquid which overflow is 20 g

The mass of the liquid which overflow = mass of the stone = 20 g

1 gram of the liquid occupies 1 cm³ of space.

20 g of the liquid will occupy;
`(20 * 1)/(1)` = 20 cm³

(a) Since the volume of the water displaced is equal to the volume of the stone.

∴ The volume of the stone = 20 cm³

(b) Mass = density × volume

Density of the stone = 5.0 g/cm³

Volume of the stone = 20 cm³

Mass of the stone = 5 g/cm³ × 20 cm³ = 100 g