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alexdok [17]
3 years ago
11

Determine the work done by an ideal gas while expanding by a volume of 0.25 L against an external pressure of 1.50 atm (assume a

diabatic conditions, i.e., no heat exchange during the expansion).
Chemistry
1 answer:
Sliva [168]3 years ago
6 0

Answer : The value of work done by an ideal gas is, 37.9 J

Explanation :

Formula used :

Expansion work = External pressure of gas × Volume  of gas

Expansion work = 1.50 atm × 0.25 L

Expansion work = 0.375 L.atm

Conversion used : (1 L.atm = 101.3 J)

Expansion work = 0.375 × 101.3 = 37.9 J

Therefore, the value of work done by an ideal gas is, 37.9 J

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3 0
3 years ago
What is the molarity (M) of the following solutions?
Dennis_Churaev [7]

Answer:

The molarity (M) of the following solutions are :

A. M = 0.88 M

B. M = 0.76 M

Explanation:

A. Molarity (M) of 19.2 g of Al(OH)3 dissolved in water to make 280 mL of solution.

Molar mass of Al(OH)3 = Mass of Al + 3(mass of O + mass of H)

                                      = 27 + 3(16 + 1)

                                      = 27 + 3(17) = 27 + 51

                                      = 78 g/mole

Al(OH)_3 = 78 g/mole

Given mass= 19.2 g/mole

Mole = \frac{Given\ mass}{Molar\ mass}

Mole = \frac{19.2}{78}

Moles = 0.246

Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}

Volume = 280 mL = 0.280 L

Molarity = \frac{0.246}{0.280)}

Molarity  = 0.879 M

Molarity  = 0.88 M

B .The molarity (M) of a 2.6 L solution made with 235.9 g of KBr​

Molar mass of KBr = 119 g/mole

Given mass = 235.9 g

Mole = \frac{235.9}{119}

Moles = 1.98

Volume = 2.6 L

Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}

Molarity = \frac{1.98}{2.6)}

Molarity = 0.762 M

Molarity = 0.76 M

4 0
3 years ago
Balance the following: ___ AlBr3+ ___ K---> ___KBr+ ___ Al
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___AlBr3 + ___K -> ___KBr + ___ Al

1 AlBr3 + 3K -> 3KBr + 1 Al

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6 0
4 years ago
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