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2 years ago

30. The density of an unknown gas at 27°C and 2 atm pressure is equal with density of N2 gas at

Chemistry

1 answer:

Zanzabum2 years ago

6 0

Answer:

Molar mass of the unknown gas is 64.6 g/mol

Explanation:

Let's think this excersise with the Ideal Gases Law.

We start from the N₂. At STP conditions we know that 1 mol of anything occupies 22.4L.

We apply: P . V = n . R . T

5 atm . V = 1 mol . 0.082 . 325K

V = (1 mol . 0.082 . 325K) / 5 atm = 5.33 L

It is reasonable to say that, if we have more pressure, we may have less volume.

As this is the volume for 1 mol of N₂, our mass is 28 g. Then, the density of the nitrogen and the unknown gas is 28 g/5.33L = 5.25 g/L

Our unknown gas has, this density at 27°C and 2 atm.

If we star from this, again: 1 mol of any gas occupy 22.4L at STP, we can calculate the volume for 1 mol at those conditions:

P₁ . V₁ / T₁ = P₂ . V₂ / T₂

1 atm . 22,4L / 273K = 2 atm . V₂ / 300K

Remember that the value for T° is Absolute (T°C + 273)

[ (1 atm . 22.4L / 273K) . 300K] / 2 atm = V₂ → 12.3L

This is the volume for 1 mol of the unknown gas at 2 atm and 27°C

We use density to determine the mass: 12.3 L . 5.25 g/L = 64.6 g

That's the molar mass: 64.6 g/mol

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The gas left in a used aerosol can is at a pressure of 103 kPa at 25.0 °C. If the can heats up to 50.0 °C, what is the pressure

Anton [14]

The question mentions a change in temperature from 25 to 50 °C. With that, the aim of the question is to determine the change in volume based on that change in temperature. Therefore this question is based on Gay- Lussac's Gas Law which notes that an increase in temperature, causes an increase in pressure since the two are directly proportional (once volume remains constant). Thus Gay-Lussac's Equation can be used to solve for the answer.
Boyle's Equation: $\frac{P_{1} }{T_{1} }$ = $\frac{P_{2} }{T_{2} }$
Since the initial temperature (T₁) is 25 C, the final temperature is 50 C (T₂) and the initial pressure (P₁) is 103 kPa, then we can substitute these into the equation to find the final pressure (P₂).
$\frac{P_{1} }{T_{1} }$ = $\frac{P_{2} }{T_{2} }$ ∴ by substituting the known values, ⇒ (103 kPa) ÷ (25 °C) = (P₂) ÷ (50 °C)
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3 years ago

If a solution has a pH of 10, is the solution acidic or basic?Explain why

natka813 [3]

Basic because a pH greater than 7 is basic

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2 years ago

ΔG o for the reaction H2(g) + I2(g) ⇌ 2HI(g) is 2.60 kJ/mol at 25°C. Calculate ΔG o , and predict the direction in which the rea

kondaur [170]

Answer:

The reaction is not spontaneous in the forward direction, but in the reverse direction.

Explanation:

Step 1: Data given

H2(g) + I2(g) ⇌ 2HI(g) ΔG° = 2.60 kJ/mol

Temperature = 25°C = 25+273 = 298 Kelvin

The initial pressures are:

pH2 = 3.10 atm

pI2 = 1.5 atm

pHI 1.75 atm

Step 2: Calculate ΔG

ΔG = ΔG° + RTln Q

with ΔG° = 2.60 kJ/mol

with R = 8.3145 J/K*mol

with T = 298 Kelvin

Q = the reaction quotient → has the same expression as equilibrium constant → in this case Kp = [p(HI)]²/ [p(H2)] [p(I2)]

with pH2 = 3.10 atm

pI2 = 1.5 atm

pHI 1.75 atm

Q = (3.10²)/(1.5*1.75)

Q = 3.661

ΔG = ΔG° + RTln Q

ΔG = 2600 J/mol + 8.3145 J/K*mol * 298 K * ln(3.661)

ΔG =5815.43 J/mol = 5.815 kJ/mol

To be spontaneous, ΔG should be <0.

ΔG >>0 so the reaction is not spontaneous in the forward direction, but in the reverse direction.

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3 years ago

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PSYCHO15rus [73]

Answer:

D.Salad

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