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larisa [96]
2 years ago
14

1. A gas has a volume of 140L at 37 ºC and 620 kpa pressure. Calculate its volume at STP.

Chemistry
1 answer:
solniwko [45]2 years ago
5 0

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%

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3 years ago
Rewrite the equation below using the correct conventions for stoichiometric coefficients. 2 BaCl2 + 2 H2SO4 → 2 BaSO4 + 4 HCl
vichka [17]

Answer:

BaCl2+H2SO4 \rightarrow BaSO4+2HCl

Explanation:

The given equation is:

2 BaCl2+2H2SO4 \rightarrow2BaSO4+4HCl

Based on the reaction stoichiometry:

2 moles of barium chloride (BaCl2) reacts with 2 moles of sulfuric acid (H2SO4) to form 2 moles of barium sulfate (BaSO4) and 4 moles of HCl

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BaCl2+H2SO4 \rightarrow BaSO4+2HCl

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3 years ago
A 5.00g of X, the product of organic synthesis is obtained in a 1.0 dm3 aqueous solution. Calculate the mass of X that can be ex
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Answer:

mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g

Explanation:

The partition coefficient of X between ethoxy ethane (ether) and water, K is given by the formula

K = concentration of X in ether/concentration of X in water

Partition coefficient, K(X) between ethoxy ethane and water = 40

Concentration of X in ether = mass(g)/volume(dm³)

Mass of X in ether = m g

Volume of ether = 50/1000 dm³ = 0.05 dm³

Concentration of X in ether = (m/0.05) g/dm³

Concentration of X in water = mass(g)/volume(dm³)

Mass of X in water left after extraction with ether = (5 - m) g

Volume of water = 1 dm³

Concentration of X in water = (5 - m/1) g/dm³

Using K = concentration of X in ether/concentration of X in water;

40 = (m/0.05)/(5 - m)

(m/0.05) = 40 × (5 - m)

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m = 0.05 × (200 - 40m)

m = 10 - 2m

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m = 3.33 g of X

Therefore, mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g

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Answer:

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