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velikii [3]
3 years ago
15

HELP PLEASE (Solve this problem using the appropriate law). what is the volume of 0.382 moles of hydrogen gas at 1.50 atmosphere

s pressure and a temperature of 295k?
(R=0.0821L•atm/mol•k)

6.17L
1.2L
3.08L
4.01L
0.23L
Chemistry
1 answer:
kaheart [24]3 years ago
5 0

Answer:

V = 6.17 L

Explanation:

Given data:

Volume = ?

Number of moles = 0.382 mol

Pressure = 1.50 atm

Temperature = 295 k

R = 0.0821 L. atm. /mol. k

Solution:

According to ideal gas equation:

PV= nRT

V = nRT/P

V = 0.382 mol × 0.0821 L. atm. /mol. k ×295 k / 1.50 atm

V = 9.252  L. atm.  / 1.50 atm

V = 6.17 L

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7 0
3 years ago
An atom of an element has 54 protons. Some of the elements atoms have 77 neutrons, while other atoms have 79 neutrons. What are
Alja [10]

Answer:

The atomic numbers of both are 54 while the mass numbers are 131 and 133 respectively.

Explanation:

- The number of protons = Atomic number. So, if the atom has 54 protons and it remained unchanged, then the two types of atoms of this element both have atomic numbers of 54.

- On the other hand, mass number is the sum of protons and neutrons

So, if type 1 has protons = 54 and neutrons = 77: mass number = 54 + 77 = 131

if Type 2 has protons = 54 and neutrons = 79: mass number = 54 + 77 = 133

(Since the possibility of atoms of the same element to have different mass numbers but the same atomic number is called isotopy). The two types of atoms with mass numbers 131 and 133 described are isotopes.

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3 years ago
The stopcock connecting a 3.06 L bulb containing methane gas at a pressure of 9.61 atm, and a 6.65 L bulb containing oxygen gas
Contact [7]

Answer : The final pressure in the system is 4.22 atm.

Explanation :

First we have to calculate the moles of methane.

PV=n_1RT

where,

P = pressure of gas = 9.61 atm

V = volume of gas = 3.06 L

T = temperature of gas = T

n_1 = number of moles of methane gas = ?

R = gas constant

Now put all the given values in the ideal gas equation, we get:

(9.61atm)\times (3.06L)=n_1\times RT

n_1=\frac{29.4}{RT}

Now we have to calculate the moles of oxygen gas.

PV=n_2RT

where,

P = pressure of gas = 1.75 atm

V = volume of gas = 6.65 L

T = temperature of gas = T

n_2 = number of moles of oxygen gas = ?

R = gas constant

Now put all the given values in the ideal gas equation, we get:

(1.75atm)\times (6.65L)=n_2\times RT

n_2=\frac{11.6}{RT}

Now we have to determine the final pressure in the system after mixing the gases.

P_{total}=(n_1+n_2)\times \frac{RT}{V_{total}}

where,

P_{total} = final pressure of gas = ?

V_{total} = final volume of gas = (3.06 + 6.65)L = 9.71 L

T = temperature of gas = T

R = gas constant

Now put all the given values in the ideal gas equation, we get:

P_{total}=(\frac{29.4}{RT}+\frac{11.6}{RT})\times \frac{RT}{9.71L}

P_{total}=4.22atm

Therefore, the final pressure in the system is 4.22 atm.

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Triss [41]

Answer:

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