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LenaWriter [7]
3 years ago
11

Which gas law states that the pressure of a gas decreases when volume is increased and the temperature is unchanged

Chemistry
2 answers:
andrey2020 [161]3 years ago
4 0
Boyle's law state that the total pressure of an enclosed gas is inversely proportional to the volume of the container, as long as the temperature is constant
Nesterboy [21]3 years ago
3 0

Answer: Boyle's law

Explanation: It states that pressure of a gas is inversely proportional to the volume of a gas at constant temperature.

P∝\frac{1}{V}       at constant temperature

or If the pressure of gas is decreased, the volume increases if the temperature is unchanged and If the pressure of gas is increased, the volume decreases if the temperature is unchanged.

P_1V_1=P_2V_2

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What is the concentration (m) of ch3oh in a solution prepared by dissolving 16.8 g of ch3oh in sufficient water to give exactly
Natalija [7]

The Molarity concentration is expressed in units of moles / L. So let us first determine the number of moles of CH3OH, and then divide that amount by the total volume of 0.230 L of solution.<span>

<span>To determine the number of moles of CH3OH, divide the weight in grams of CH3OH by the molecular weight of CH3OH: (MW of CH3OH = 32 g / mol)</span>

number of moles = 16.8 g / (32 g / mol)</span>

number of moles = 0.525 mol CH3OH <span>

Then we calculate for molarity:</span>

Molarity = 0.525 mol CH3OH / .230 L

Molarity = 2.2826 mol / L

<span>Molarity = 2.28 M</span>

5 0
3 years ago
The reaction of 2h2+o2--&gt;2H2o + heat is
GenaCL600 [577]
That is an exothermic reaction since it releases heat with the product.
6 0
3 years ago
Read 2 more answers
A 115.0-g sample of oxygen was produced by heating 400.0 g of potassium chlorate.2KClO3 Right arrow. 2KCI + 3O2What is the perce
STALIN [3.7K]

Answer:

73.4% is the percent yield

Explanation:

2KClO₃ →  2KCl  + 3O₂

This is a decomposition reaction, where 2 moles of potassium chlorate decompose to 2 moles of potassium chloride and 3 moles of oxygen.

We determine the moles of salt: 400 g . 1. mol /122.5g= 3.26 moles of KClO₃

In the theoretical yield of the reaction we say:

2 moles of potassium chlorate can produce 3 moles of oxygen

Therefore, 3.26 moles of salt, may produce (3.26 . 3) /2 = 4.89 moles of O₂

The mass of produced oxygen is: 4.89 mol . 32 g /1mol = 156.6g

But, we have produced 115 g. Let's determine the percent yield of reaction

Percent yield = (Produced yield/Theoretical yield) . 100

(115g / 156.6g) . 100 = 73.4 %

5 0
3 years ago
If 43.1 g of O2 and 6.8 g of CO2 are placed in a 13.7 L container at 34 degrees C, what is the mixture of gasses?
jonny [76]

Answer:

The total pressure of the gas mixture = 2.76 atm

Note: The question is not complete. The complete question is as follow:

If 43.1 g of O2 and 6.8 g of CO2 are placed in a 13.7 L container at 34 degree Celsius , what is the pressure of the mixture of gases?

Explanation:

Mass of O₂ gas = 43.1 g, molar mass of O₂ gas = 32.0 g/mol

Number of moles of O₂ gas = 43.1/32.0 = 1.347 moles

Mass of CO₂ gas = 6.8 g, molar mass of CO₂ gas = 44.0 g

Number of moles of CO₂ gas = 6.8/44 = 0.155 moles

Total number of moles of gas mixture, n = (1.347 + 0.155) = 1.502 moles

Volume of gas mixture, V = 13.7 L

Temperature of gas mixture, T = 34 °C = (273.15 + 34) K = 307.15 K

Pressure of gas mixture = ?

Molar gas constant, R = 0.0821 liter·atm/mol·K.

Using the ideal gas equation: PV =nRT

P = nRT/V

P = (1.502 × 0.0821 × 307.15) / 13.7

P = 2.76 atm

Therefore, the total pressure of the gas mixture = 2.76 atm

5 0
3 years ago
A molecular biologist measures the mass of cofactor a in an average yeast cell. the mass is 41.5 pg . what is the total mass in
Hitman42 [59]
The unit pg stands for pictogram. It is one-trillionth of a gram. Because of the very small mass, it is expressed in the prefix form of the base units for convenience. Now, the mass of cofactor a is 41.5 pg per cell. Since there are a total of 105 cells, the total mass would be:

Total mass = 105 cells * 41.5 pg/cell = 4,357.5 pg 
6 0
3 years ago
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