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2 years ago

If the density of the gas is 1.2 g/L at 745. torr and 20 C, what is its molecular mass?

Chemistry

1 answer:

tigry1 [53]2 years ago

7 0

Explanation:

If you take the ideal gas law of PV = nRT and modify it to

P*molar mass = density*RT

(745/760)*M = 1.2*0.08206*293

Solve for M = molar mass.

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What is the mass in grams of 0.0015 mol of aspirin (C9H8O4)

almond37 [142]

Answer:

116.1 g

Explanation:

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2 years ago

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A sample of carbon dioxide at RTP is 0.50 dm3. How many grams of carbon dioxide do we have?

prohojiy [21]

Answer:

0.924 g

Explanation:

The following data were obtained from the question:

Volume of CO2 at RTP = 0.50 dm³

Mass of CO2 =?

Next, we shall determine the number of mole of CO2 that occupied 0.50 dm³ at RTP (room temperature and pressure). This can be obtained as follow:

1 mole of gas = 24 dm³ at RTP

Thus,

1 mole of CO2 occupies 24 dm³ at RTP.

Therefore, Xmol of CO2 will occupy 0.50 dm³ at RTP i.e

Xmol of CO2 = 0.5 /24

Xmol of CO2 = 0.021 mole

Thus, 0.021 mole of CO2 occupied 0.5 dm³ at RTP.

Finally, we shall determine the mass of CO2 as follow:

Mole of CO2 = 0.021 mole

Molar mass of CO2 = 12 + (2×16) = 13 + 32 = 44 g/mol

Mass of CO2 =?

Mole = mass /Molar mass

0.021 = mass of CO2 /44

Cross multiply

Mass of CO2 = 0.021 × 44

Mass of CO2 = 0.924 g.

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3 years ago

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dlinn [17]

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3 years ago

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The volume (in L) that would be occupied by 5.00 mols of 02 at STP is

crimeas [40]

Answer : The volume of oxygen at STP is 112.0665 L

Solution : Given,

The number of moles of $O_2$ = 5 moles

At STP, the temperature is 273 K and pressure is 1 atm.

Using ideal gas law equation :

$PV=nRT$

where,

P = pressure of gas

V = volume of gas

n = the number of moles

T = temperature of gas

R = gas constant = 0.0821 L atm/mole K (Given)

By rearranging the above ideal gas law equation, we get

$V=\frac{nRT}{P}$

Now put all the given values in this expression, we get the value of volume.

$V=\frac{(5moles)\times (0.0821Latm/moleK)\times (273K)}{1atm}=112.0665L$

Therefore, the volume of oxygen at STP is 112.0665 L

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3 years ago

If 5.57 g of Ag2O is sealed in a 75.0-mL tube filled with 760 torr of N2 gas at 28 ∘C, and the tube is heated to 310 ∘C, the Ag2

Snezhnost [94]

Answer: Total pressure = 7293.2 torr or 9.60 atm

Explanation:

<em>Total pressure = partial pressure of nitrogen + partial pressure of oxygen</em>

The partial pressure due to nitrogen is determined using the equation of Gay-Lussac's law: <em>P₁/T₁=P₂/T₂</em>

P₁ = 760 torr = 1atm, T₁ = 28∘C = (273+28)K = 301k, P₂ = ?, T₂ = 310∘C =(310+273)K = 583K

P₂ = P₁ T₂/ T₁

P₂ = 760 * 583 / 301 = 1472.03 torr

The pressure due to Oxygen gas produced is calculated thus:

Balanced equation of the decomposition of Ag₂O at s.t.p. is as follows;

2Ag₂O ----> 4Ag + O₂(g)

2 moles of Ag₂O produces 1 mole of O₂

molar mass of Ag₂O = (2*108 + 16)g = 232g/mol

molar volume of gas at s.t.p. = 22.4L

2*232g i.e. 464g of Ag₂O produces 22.4L of O₂

5.57g of Ag₂O will produce 5.57g*22.4L/464g = 0.269L or 269mL of O₂

Using the General gas equation P₁V₁/T₁=P₂V₂/T₂

P₁ = 1atm = 760 torr, V₁ = 269mL, T₁=273K, P₂ = ?, V₂= 75mL, T₂ = 583K

P₂ = P₁V₁T₂/V₂T₁

P₂ = 760*269*583 / 75*273

P₂ = 5821.17 torr

Total pressure = (1472.03 + 5821.17) torr

Total pressure = 7293.2 torr or 9.60 atm

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3 years ago