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snow_tiger [21]
3 years ago
6

A 8.2 L sample of gas has a pressure of 0.8 atm at a temperature of 259 K. If the temperature increases to 301 K, causing the vo

lume to increase to 11.5 L, what is the new pressure? Round your answer to the nearest tenth
Chemistry
2 answers:
Rina8888 [55]3 years ago
8 0

The answer is 0.7 atm


pashok25 [27]3 years ago
8 0

Answer:

0.7 atm

Explanation:

For the Clapeyron's equation, we can calculate the change in the physical properties that happens when there is a state change:

P1*V1/T1 = P2*V2/T2

Where P is the pressure, V is the volume, T is the temperature, 1 is the initial state, and 2 the final state. So:

0.8*8.2/259 = P2*11.5/301

0.02533 = 0.038206*P2

P2 = 0.7 atm

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A metal (M) forms an oxide with the formula MO. If the oxide contains 7.17 % O by mass, what is the identity of the metal?
Nataly_w [17]

The metal is lead, Pb.

One unit of the oxide contains one atom of O (16.00 u).

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3 years ago
Plz help me
AlekseyPX
All of the above would be the answer


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2 years ago
Read 2 more answers
A 3.42 gram sample of an unknown gas is found to occupy a volume of 1.90 L at a pressure of 547 mm Hg and a temperature of 33 °C
liq [111]
Use the Ideal Gas Law to find the moles of gas first.

Be sure to convert T from Celsius to Kelvin by adding 273.

Also I prefer to deal with pressure in atm rather than mmHg, so divide the pressure by 760 to get it in atm.

PV = nRT —> n = PV/RT
P = 547 mmHg = 547/760 atm = 0.720 atm
V = 1.90 L
T = 33°C = 33 + 273 K = 306 K
R = 0.08206 L atm / mol K

n = (0.720 atm)(1.90 L) / (0.08206 L atm / mol K)(306 K) = 0.0545 mol of gas

Now divide grams by mol to get the molecular weight.

3.42 g / 0.0545 mol = 62.8 g/mol
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