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mezya [45]
3 years ago
10

A sample of gas X occupies 10 mº at a pressure of 120 kPa.

Chemistry
1 answer:
Mashutka [201]3 years ago
5 0

Answer:

The new pressure of the gas comes out to be 400 KPa.

Explanation:

Initial volume of gas = V = 10\textrm{ m}^{3}

Initial pressure of gas = P = 120 KPa

Final volume of gas = V' = 3\textrm{ m}^{3}

Assuming temperature to be kept constant.

Assuming final pressure of the gas to be P' KPa

PV = P'V' \\120\textrm{ KPa}\times 10\textrm{ m}^{3} = \textrm{P'}\times 3\textrm{ m}^{3} \\\textrm{P'} = 400\textrm{ KPa}

New pressure of gas = 400 KPa

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Does the antifreeze you put in a car radiator have a lower or higher freezing point than water
saveliy_v [14]
Hello!
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7 0
3 years ago
A gas is collected at 20.0 °C and 725.0 mm Hg. When the temperature is
krek1111 [17]

Answer:

676mmHg

Explanation:

Using the formula;

P1/T1 = P2/T2

Where;

P1 = initial pressure (mmHg)

P2 = final pressure (mmHg)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question;

P1 = 725.0mmHg

P2 = ?

T1 = 20°C = 20 + 273 = 293K

T2 = 0°C = 0 + 273 = 273K

Using P1/T1 = P2/T2

725/293 = P2/273

Cross multiply

725 × 273 = 293 × P2

197925 = 293P2

P2 = 197925 ÷ 293

P2 = 676mmHg.

The resulting pressure is 676mmHg

3 0
3 years ago
Can you solve question 5 and 6 only​
Yanka [14]

Answer:

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4 0
3 years ago
If you react 156.0 g of calcium chloride with an excess NaOH, how much sodium chloride should you get?
laiz [17]

Answer:

164.3g of NaCl

Explanation:

Based on the chemical equation:

CaCl2 + 2NaOH → 2NaCl + Ca(OH)2

<em>where 1 mole of CaCl2 reacts with 2 moles of NaOH</em>

To solve this question we must convert the mass of CaCl2 to moles. Using the chemical equation we can find the moles of NaCl and its mass:

<em>Moles CaCl2 -Molar mass: 110.98g/mol-</em>

156.0g CaCl₂ * (1mol / 110.98g) = 1.4057 moles CaCl2

<em>Moles NaCl:</em>

1.4057 moles CaCl2 * (2mol NaCl / 1mol CaCl2) = 2.811 moles NaCl

<em>Mass NaCl -Molar mass: 58.44g/mol-</em>

2.811 moles NaCl * (58.44g / mol) = 164.3g of NaCl

7 0
3 years ago
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