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4 years ago

Explain why liquids unlike gases are virtually incompressible

Chemistry

1 answer:

MaRussiya [10]4 years ago

8 0

Lipids cannot be compressed since there is only a small distance between the molecules when bonded

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3 years ago

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3 years ago

Fractional distilation of liquid air usually produces nitrogen and oxygen as major products . Substance used to remove carbon iv

Nastasia [14]

Answer:

Caustic soda

Explanation:

The fractional distillation of air is carried out on liquid air. Before air is liquified, the carbon dioxide content of air is removed using caustic soda. The air is then compressed to a pressure of about 200 atm, sudden expansion of the gas leads to cooling. The process continues until air becomes liquid at -200°C.

Fractional distillation of liquid air usually produces nitrogen and oxygen as major products. nitrogen in obtained first since it has a lower boiling point than oxygen. The gases are then dried, compressed and stored in cylinders.

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3 years ago

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3 years ago

Read 2 more answers

If 11.7 g of aluminum reacts with 37.2 g of copper (II) sulfate according to the following reaction, how many grams of aluminum

zhannawk [14.2K]

Answer:

There is 26.59 grams of aluminium sulfate produced

Explanation:

Step 1: Data given

Mass of aluminium = 11.7 grams

Mass of copper (II) sulfate = 37.2 grams

Molar mass of Aluminium = 26.98 g/mol

Molar mass of CuSO4 = 159.61 g/mol

Molar mass of Al2(SO4)3 = 342.15 g/mol

Step 2: The balanced equation

2Al + 3CuSO4 → Al2(SO4)3 + 3Cu

Step 3: Calculate moles of Aluminium

Moles Al = mass Al / molar mass Al

Moles Al = 11.7 grams / 26.98 g/mol

Moles Al = 0.434 mol

Step 4: Calculate moles of CuSO4

Moles CuSO4 = 37.2 grams / 159.61 g/mol

Moles CuSO4 = 0.233 moles

Step 5: Calculate limiting reactant

For 2 moles of Al we need 3 moles of CuSO4

CuSO4 is the limiting reactant. It will be completely consumed (0.233 moles).

Al is in excess. There will be consumed 0.233 *(2/3) = 0.1553 moles

There will remain 0.434 - 0.1553 = 0.2787 moles

Step 6: Calculate moles of Al2(SO4)3

For 2 moles of Al we need 3 moles of CuSO4, to produce 1 mole of Al2(SO4)3 and 3 moles of Cu

For 0.233 moles CuSO4 we produce 0.233/3 = 0.0777 moles of Al2(SO4)3

Step 7: Calculate mass of Al2(SO4)3

Mass of Al2(SO4)3 = moles Al2(SO4)3 * molar mass Al2(SO4)3

Mass of Al2(SO4)3 = 0.0777 moles * 342.15g/mol

Mass of Al2(SO4)3 = 26.59 grams

There is 26.59 grams of aluminium sulfate produced

4 0

3 years ago