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Mrrafil [7]
3 years ago
10

A sample of gas is kept at a temperature of 300 k, a volume of 1.45L, and a pressure of 3.00 kPa. How many moles of has are in t

he sample
Chemistry
1 answer:
klasskru [66]3 years ago
7 0

Answer:

the number of moles of the gas is 0.00174 mol.

Explanation:

Given;

temperature of the, T = 300 K

volume of the gas, V = 1.45 L

Pressure of the gas, P = 3.0 kPa

The number of moles of the gas is calculated using Ideal gas equation, as follows;

PV = nRT

Where;

n is number of moles of the gas

R is the ideal gas constant, = 8.315 L.kPa/mol.K

n = PV / RT

n = (3 x 1.45) / (8.315 x 300)

n = 0.00174 mol.

Therefore, the number of moles of the gas is 0.00174 mol.

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5 0
3 years ago
The volume occupied by 1.5 mole of gas at 35°C and 2.0 atmosphere of pressure is blank liters
Aneli [31]

Answer: 18.97 L

Explanation:

This can be solved by the Ideal Gas equation:

P.V=n.R.T  

Where:  

P=2 atm is the pressure of the gas  

V is the volume of the gas  

n=1.5 mole the number of moles of gas  

R=0.0821\frac{L.atm}{mol.K} is the gas constant  

T=35\°C+273.15=308.15 K is the absolute temperature of the gas in Kelvin

Finding V:

V=\frac{nRT}{P}

V=\frac{(1.5 mole)(0.0821\frac{L.atm}{mol.K})(308.15 K)}{2 atm}

V=18.97 L

Therefore:

The volume occupied by 1.5 mole of gas at 35°C and 2.0 atmosphere of pressure is <u>18.97</u> liters

7 0
4 years ago
What is the molarity of 2.5 liters of solution containing 5 moles of NaCl?
Kryger [21]

Answer:

2 mol/L

Explanation:

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5 0
2 years ago
The gas phase reaction 2 N2O5(g) → 4 NO2(g) + O2(g) has an activation energy of 103 kJ/mol, and the first order rate constant is
vaieri [72.5K]

Answer:  The rate constant at 296 K is 1.51\times 10^{-3}min^{-1}

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

or,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 266 K = 1.35\times 10^{-5}min^{-1}

K_2 = rate constant at 296 K = ?

Ea = activation energy for the reaction = 103 kJ/mol = 103000 J/mol

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 266 K

T_2 = final temperature = 296 K

Now put all the given values in this formula, we get

\log (\frac{K_2}{1.35\times 10^{-5}})=\frac{103000}{2.303\times 8.314J/mole.K}[\frac{1}{266}-\frac{1}{296}]

\log (\frac{K_2}{1.35\times 10^{-5}})=2.049

K_2=1.51\times 10^{-3}min^{-1}

Thus the rate constant at 296 K is 1.51\times 10^{-3}min^{-1}

3 0
3 years ago
Explain the impact of water cycle to humans. I couldnt find the subject science so it science
allsm [11]

Answer:

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Explanation:

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