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photoshop1234 [79]
3 years ago
11

Pushing down on the piston causes an increase in pressure and a decrease in the volume of the gas in the cylinder. This is BEST

explained by the fact that
Chemistry
2 answers:
Mamont248 [21]3 years ago
6 0

Answer: The answer is D on usa test prep

Explanation:The decrease in volume as a result in an increase in pressure is BEST explained by the fact that there is a large relative distance between gas molecules. This explains the property of compressibility, exhibited by gases under pressure.

qaws [65]3 years ago
3 0

Pushing down on the piston causes an increase in pressure and a decrease in the volume of the gas in the cylinder. This is BEST explained by the fact that the pressure is always inversely proportional with volume at constant temperature and number of moles of substance.

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94%

Explanation:

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3 years ago
A sulfuric acid solution containing 571.3 g of h2so4 per liter of aqueous solution has a density of 1.329 g/cm3. Part a calculat
loris [4]

Mass percentage of a solution is the amount of solute present in 100 g of the solution.

Given data:

Mass of solute H2SO4 = 571.3 g

Volume of the solution = 1 lit = 1000 ml

Density of solution = 1.329 g/cm3 = 1.329 g/ml

Calculations:

Mass of the given volume of solution = 1.329 g * 1000 ml/1 ml = 1329 g

Therefore we have:

571.3 g of H2SO4 in 1329 g of the solution

Hence, the amount of H2SO4 in 100 g of solution= 571.3 *100/1329 = 42.987

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3 0
3 years ago
The following question appears on a quiz: ""You fill a tank with gas at 60°C to 100 kPa and seal it. You decrease the temperatu
dusya [7]

Answer: The final pressure will decrease ad the value is 85 kPa

Explanation:

To calculate the final pressure of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

where,

P_1\text{ and }T_1 are the initial pressure and temperature of the gas.

P_2\text{ and }T_2 are the final pressure and temperature of the gas.

We are given:

P_1=100kPa\\T_1=60^0C=(60+273)K=333K\\P_2=?\\T_2=10^0C=(10+273)K=283K

Putting values in above equation, we get:

\frac{100kPa}{333K}=\frac{P_2}{283K}\\\\P_2=85kPa

Hence, the final pressure will decrease ad the value is 85 kPa

8 0
3 years ago
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