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Dafna11 [192]
3 years ago
12

A sample of gas at 301 Kelvin and 1.5 atmospheres has a density of 3.90 g/L. What is the molar mass of this gas? Show all of the

work used to solve this problem.
Chemistry
2 answers:
Vera_Pavlovna [14]3 years ago
5 0

Answer:

Molar mass of the gas is 64.3 g/mol

Explanation:

<u>Given:</u>

Temperature of gas, T = 301 K

Pressure of gas, P = 1.5 atm

Density of gas, D = 3.90 g/L

<u>To determine:</u>

The molar mass, M of the gas

<u>Explanation</u>

Based on the ideal gas equation

PV = nRT

where P = pressure, V = volume ; R = gas constant, T = temperature

n = moles of the gas

n = \frac{mass(m)}{molar\ mass(M)}

substituting for n in the ideal gas equation:

PV = \frac{m}{M} *RT\\\\M = (\frac{m}{V})*\frac{RT}{P}  \\\\density , d = \frac{mass(m)}{Volume(V)} \\\\Therefore:\\\\M = d*\frac{RT}{P} = 3.90 g/L *\frac{.0821Latm/mol-K*301K}{1.5atm} =64.3 g/mol

meriva3 years ago
4 0
PV = nRTP is pressure, V is volume in L, n is number of moles, R is the gas constant,and T is temperature in K

(1.5 atm)(1 L) = (n)(.08206)(301K)

n = .06 moles in one liter

If there are 3.9 grams in .06 moles then

1/.06 x 3.9 = 64.2 grams per mol


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9966 [12]

Answer:

I`m pretty sure it`s electrical tape.

Explanation:

8 0
3 years ago
Many common liquids have boiling points that are less than 110°C, whereas most metals are solids at room temperature and have mu
Semmy [17]

Answer:

1. 315.4 K

2. 1560 °C

Explanation:

To convert from celsius to Kelvin, the following formula can be used:

T(K) = T(°C) + 273

Where:

T(K) => Temperature in Kelvin

T(°C) => Temperature in degree celsius

1. Determination of the temperature in Kelvin.

Temperature (T) in °C = 42.4 °C.

Temperature (T) in K =?

T(K) = T(°C) + 273

T(K) = 42.4 °C + 273

T(K) = 315.4 K

2. Determination of the temperature in degree Celsius.

Temperature (T) in K = 1833 K

Temperature (T) in °C =?

T(K) = T(°C) + 273

1833 = T(°C) + 273

Collect like terms

T(°C) = 1833 – 273

T(°C) = 1560 °C

7 0
3 years ago
What is the resulting formula unit when magnesium and nitrogen bond
muminat
The answer is Mg3N2.
5 0
3 years ago
Read 2 more answers
What volume of nitrogen gas at STP would react with 37.2 g of magnesium to produce magnesium nitride
anastassius [24]

Answer:

11.58 L of N₂

Explanation:

We'll begin by calculating the number of mole in 37.2 g of magnesium. This can be obtained as follow:

Mass of Mg = 37.2 g

Molar mass of Mg = 24 g/mol

Mole of Mg =?

Mole = mass /Molar mass

Mole of Mg = 37.2 / 24

Mole of Mg = 1.55 moles

Next, we shall write the balanced equation for the reaction. This is illustrated below:

3Mg + N₂ —> Mg₃N₂

From the balanced equation above,

3 moles of Mg reacted with 1 mole of N₂.

Therefore, 1.55 moles of Mg will react with = (1.55 × 1)/3 = 0.517 mole of N₂

Thus, 0.517 mole of N₂ is need for the reaction.

Finally, we shall determine the volume of N₂ needed for the reaction as follow:

Recall:

1 mole of a gas occupies 22.4 L at STP.

1 mole of N₂ occupied 22.4 L at STP.

Therefore, 0.517 mole of N₂ will occupy = 0.517 × 22.4 = 11.58 L at STP

Thus, 11.58 L of N₂ is needed for the reaction.

6 0
3 years ago
If two gases, A and B, in separate 1 liter containers exert
babunello [35]

Answer:

5Atm

Explanation:

I just guess and it’s right

5 0
3 years ago
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