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3 years ago

A sample of 35.1 g of methane gas has a volume of 5.20 l at a pressure of 2.70 atm. calculate the temperature of the gas.

1 answer:

6 0

We can use the ideal gas equation to determine the temperature with the given conditions of mass of the gas, volume, and pressure. The equation is expressed
PV=nRT where n is the number of moles equal to mass / molar mass of gas. Substituting the given conditions with R = 0.0521 L atm/mol K we can find the temperature

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How many moles of sulfur trioxide are formed from 3 moles of oxygen using the given

fiasKO [112]

Answer:

6 moles of SO₃ formed.

Explanation:

Given data:

Number of moles of SO₃ formed = ?

Number of moles of oxygen react = 3 mol

Solution:

Chemical equation;

2SO₂+ O₂ → 2SO₃

now we will compare the moles of oxygen and sulfur trioxide.

O₂ : SO₃

1 : 2

3 : 2/1×3 = 6 moles

Thus, six moles of SO₃ will formed.

8 0

3 years ago

There are two common oxides of copper; one is 80% copper and the other is 89% copper by weight. Calculate the formula and name e

irakobra [83]

80% copper (Cu)

Cu: 80 : 63.546 = 1.259

O: 20 : 16 = 1.25

Cu:O = 1 : 1

the formula: CuO

89% copper (Cu)

Cu: 89 : 63.546 = 1.4

O: 11 : 16 = 0.6875

Cu:O = 2:1

the formula: Cu₂O

8 0

2 years ago

A student weighs an empty flask and stopper and finds the mass to be 55.844 g. She then adds about 5 mL of an unknown liquid and

Oduvanchick [21]

Answer :

(a) The pressure of the vapor in the flask in atm is, 0.989 atm

(b) The temperature of the vapor in the flask in Kelvin is, 372.7 K

The volume of the flask in liters is, 0.2481 L

(c) The mass of vapor present in the flask was, 0.257 g

(d) The number of moles of vapor present are 0.00802 mole.

(e) The mass of one mole of vapor is 32.0 g/mole

Explanation : Given,

Mass of empty flask and stopper = 55.844 g

Volume of liquid = 5 mL

Temperature = $99.7^oC$

Mass of flask and condensed vapor = 56.101 g

Volume of flask = 248.1 mL

Barometric pressure in the laboratory = 752 mmHg

(a) First we have to determine the pressure of the vapor in the flask in atm.

Pressure of the vapor in the flask = Barometric pressure in the laboratory = 752 mmHg

Conversion used :

$1atm=760mmHg$

or,

$1mmHg=\frac{1}{760}atm$

As, $1mmHg=\frac{1}{760}atm$

So, $752mmHg=\frac{752mmHg}{1mmHg}\times \frac{1}{760}atm=0.989atm$

Thus, the pressure of the vapor in the flask in atm is, 0.989 atm

(b) Now we have to determine the temperature of the vapor in the flask in Kelvin.

Conversion used :

$K=273+^oC$

As, $K=273+^oC$

So, $K=273+99.7=372.7$

Thus, the temperature of the vapor in the flask in Kelvin is, 372.7 K

Now we have to determine the volume of the flask in liters.

Conversion used :

1 L = 1000 mL

or,

1 mL = 0.001 L

As, 1 mL = 0.001 L

So, 248.1 mL = 248.1 × 0.001 L = 0.2481 L

Thus, the volume of the flask in liters is, 0.2481 L

(c) Now we have to determine the mass of vapor that was present in the flask.

Mass of flask and condensed vapor = 56.101 g

Mass of empty flask and stopper = 55.844 g

Mass of vapor in flask = Mass of flask and condensed vapor - Mass of empty flask and stopper

Mass of vapor in flask = 56.101 g - 55.844 g

Mass of vapor in flask = 0.257 g

Thus, the mass of vapor present in the flask was, 0.257 g

(d) Now we have to determine the number of moles of vapor present.

Using ideal gas equation:

PV = nRT

where,

P = Pressure of vapor = 0.989 atm

V = Volume of vapor = 0.2481 L

n = number of moles of vapor = ?

R = Gas constant = 0.0821 L.atm/mol.K

T = Temperature of vapor = 372.7 K

Putting values in above equation, we get:

$(0.989atm)\times 0.2481L=n\times (0.0821L.atm/mol.K)\times 372.7K\\\\n=0.00802mole$

Thus, the number of moles of vapor present are 0.00802 mole.

(e) Now we have to determine the mass of one mole of vapor.

$\text{Mass of one mole of vapor}=\frac{\text{Mass of vapor}}{\text{Moles of vapor}}$

$\text{Mass of one mole of vapor}=\frac{0.257g}{0.00802mole}=32.0g/mole$

Thus, the mass of one mole of vapor is 32.0 g/mole

8 0

3 years ago

A 0.04358 g sample of gas occupies 10.0-ml at 292.0 k and 1.10 atm. upon further analysis, the compound is found to be 25.305% c

kirill [66]

<span>Answer: A 0.04403 g sample of gas occupies 10.0-mL at 289.0 K and 1.10 atm. Upon further analysis, the compound is found to be 25.305% C and 74.695% Cl. What is the molecular formula of the compound? --------------------------------------... Seems like I did a problem very similar to this--this must be the "B" test. But the halogen was different. 25.305% C/12 = 2.108 74.695% Cl/35.5 = 2.104 So the empirical formula would be CH. However, there are many compounds which fit this bill, so we have to use the gas data. (And I made, in the previous problem, the simplifying assumption that 289C and 1.10 atm would offset each other, so I'll do that, too.) 0.044 grams/10 ml = x/22.4 liters 0.044g/0.010 liters = x/22.4 liters 22.4 liters/0.010 liters = 2240 (ratio) 2240 x .044 = 98.56 (actual atomic weight) CCl = 35.5+12 or 47.5, so two of those is 95 grams/mole. This is sufficiient to distinguish C2CL2, (dichloroacetylene) from C6CL6 (hexachlorobenzene) which would mass 3 times as much.</span>

3 0

3 years ago

Read 2 more answers

The product of alpha decay of the Nobelium isotope 259No is

Svet_ta [14]

Answer:

Nobelium is made by the bombardment of curium (Cm) with carbon nuclei. Its most stable isotope, 259No, has a half-life of 58 minutes and decays to Fermium (255Fm) through alpha decay or to Mendelevium (259Md) through electron capture.

Explanation:

7 0

3 years ago