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Rudiy27
3 years ago
14

Suppose you have a solid that looks like gold but you believe it to be fool’s gold. The mass of the solid is 23.5 grams. When th

e solid is lowered into a graduated cylinder the water level rises from 47.5 to 52.2 mL. Is the simple fool’s gold
Chemistry
1 answer:
frutty [35]3 years ago
3 0

Answer:

The sample is fool's gold

Explanation:

Density is defined as the ratio between mass in grams and volume in mililiters.

A sample of pure gold has a density of 19.3g/mL.

Using Archimedes' principle, the volume of the sample is:

52.2mL - 47.5mL = 4.7mL

As the mass of the sample is 23.5g, the density is:

23.5g / 4.7mL = 5g/mL

The denisty of the sample is very different to density of pure gold, that means:

<em>the sample is fool's gold</em>

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The valency of calcium is2 ? what does it mean​
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1. A gas has a volume of 140L at 37 ºC and 620 kpa pressure. Calculate its volume at STP.
solniwko [45]

Answer:

1. Volume as STP = 755 L

2. Outside temperature = 255 K

3. Percentage yield = 70.5%

Explanation:

1. At STP, pressure  = 101.3 kpa, temperature  = 0°C or 273.15 K

Using the general gas equation :

P1V1/T1 = P2V2/T2

P1 = 620 kpa

V1 = 140 L

T1 = 37°C or (273.15 + 37) K = 310.15 K

P2 = 101.3 kpa

V2 = ?

T2 = 273.15 K

V2 = P1V1T2/P2T1

V2 = 620 × 140 × 273.15 / 101.3 × 310.15

V2 = 755 L

2. Using Charles' gas law:

V1/T1 = V2/T2

V1 = 2.5 L

T1 = 290 K

V2 = 2.2 L

T2 = ?

T2 = V2T1/VI

T2 = 2.2 × 290 / 2.5

T2 = 255 K

3. Equation of reaction : 2 Al + 3 CuSO4 ---> Al2 (SO4)3 + 3 Cu

From equation of the reaction,  2 moles of Al produces 3 moles of Cu

Molar mass of Al = 27 g; Molar mass of Cu = 63.5 g

2 moles of Al = 2 × 27 g = 54 g; 3 moles of Cu = 3× 63.5 = 190.5 g

54 g of Al produces 190.5 g of Cu

1.87 g of Al will produce 190.5/54 × 1.87 g of Cu = 6.60 g of Cu

Percentage yield = actual yield /theoretical yield × 100%

Percentage yield = 4.65/6.60 × 100%

Percentage yield = 70.5%

5 0
3 years ago
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