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HACTEHA [7]
3 years ago
9

Salt formed by the partial neutralization of an acid by a base are

Chemistry
1 answer:
Makovka662 [10]3 years ago
4 0

Answer:

D. Acidic salt

Explanation:

Acidic salts:

Salts formed by incomplete neutralisation of poly-basic acids are called acidic salts. Such salts still contain one or more replaceable hydrogen atoms. These salts when neutralised by bases form normal salts.

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A containing vessel holds a gaseous mixture of nitrogen and butane. Thepressure in the vessel at 126.9 Cis 3.0 atm. At 0 C, the
Viefleur [7K]

A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.

A vessel contains a gaseous mixture of nitrogen and butane. At 126.9 °C (400.1 K) the pressure is due to the mixture is 3.0 atm.

We can calculate the total number of moles using the ideal gas equation.

P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{3.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.091 mol/L \times V

At 0 °C (273.15 K), the pressure due to the gaseous nitrogen is 1.0 atm.

We can calculate the moles of nitrogen using the ideal gas equation.

P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{1.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.030 mol/L \times V

The mole fraction of nitrogen in the mixture is:

X(N_2) = \frac{0.030 mol/L \times V}{0.091 mol/L \times V} = 0.33

A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.

Learn more: brainly.com/question/2060778

5 0
3 years ago
When a gas is cooled at constant pressure, what happens to its molecules and volume?
Dima020 [189]
Assuming it's a perfect gas, we have PV=nRT hence if T goes down, V goes down up. The volume will decrease.
3 0
3 years ago
When you apply 1000 joules of energy to 50 grams of water its temperature changes to 30 degrees . What was the initial temperatu
expeople1 [14]

Answer:

25.2°C

Explanation:

Given parameters:

Energy applied to the water  = 1000J

Mass of water  = 50g

Final temperature  = 30°C

Unknown:

Initial temperature  = ?

Solution:

To solve this problem, we use the expression below:

            H  = m c Ф

H is the energy absorbed

m is the mass

c is the specific heat capacity

Ф is the change in temperature

     1000  = 50 x 4.184 x (30  -  initial temperature )

     1000  = 209.2(30 - initial temperature)

      4.78  = 30 - initial temperature

      4.78  - 30  = - initial temperature

         Initial temperature  = 25.2°C

7 0
3 years ago
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