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Rudik [331]
2 years ago
9

A student weighs an empty flask and stopper and finds the mass to be 53.256 g. She then adds about 5 mL of an unknown liquid and

heats the flask in a boiling water bath at 98.8 degree C. After all the liquid is vaporized, she removes the flask from the bath, stoppers it, and lets it cool. After it is cool, she momentarily removes the stopper, then replaces it and weighs the flask and condensed vapor, obtaining a mass of 53.780 g. The volume of the flask is known to be 231.1 mL. The absolute atmospheric pressure in the laboratory that day is 728 mm Hg.
a. What was the pressure of the vapor in the flask in atm?
b. What was the temperature of the vapor in K? The volume of the flask in liters?
c. What was the mass of condensed vapor that was present in the flask?
d. How many moles of condensed vapor were present?
e. What is the mass of one mole of vapor?
Chemistry
1 answer:
Maru [420]2 years ago
5 0

Answer:

a) 0.957895 atm

b) 372.05 K

c) 0.524 grams

d)  0.00725 moles

e) Molar mass gas = 30.54 g/mol

Explanation:

Step 1: Data given

Mass of the flask = 53.256 grams

Volume = 5 mL

Temperature of water = 98.9 °C = 372.05 K

Mass obtained = 53.78 grams

The volume of the flask is known to be 231.1 mL = 0.2311 L

The absolute atmospheric pressure in the laboratory that day is 728 mm Hg.

Step 2: Calculate the mass of the gas

Mass gas = 53.780 - 53.256 = 0.524 grams

Step 3: Calculate moles

p*V = n*R*T

⇒with p = the pressure = 0.957895 atm

⇒with V = the volume = 0.2311 L

⇒with n = the number of moles gas = TO BE DETERMINED

⇒with R = the gas constant = 0.08206 L*atm/mol * K

⇒with T = the temperature = 372.05 K

n = (p*V)/(R*T)

n= (0.957895  * 0.2311) / (0.08206 * 372.05)

n = 0.00725 moles

Step 4: Calculate mass of the gas

Molar mass = mass / moles

Molar mass gas = 0.524 grams . 0.00725 moles

Molar mass gas = 30.54 g/mol

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A 2.80 g sample of Al reacts with 4.15 g sample of Cl2 according to the equation shown below.
solong [7]

Answer:

Mass = 5.33 g

Explanation:

Given data:

Mass of Al = 2.80 g

Mass of Cl₂ = 4.15 g

Theoretical yield of AlCl₃ = ?

Solution:

Chemical equation:

2Al  +  3Cl₂        →       2AlCl₃

Number of moles of Al:

Number of moles = mass/molar mass

Number of moles = 2.80 g/ 27 g/mol

Number of moles = 0.10 mol

Number of moles of Cl₂:

Number of moles = mass/molar mass

Number of moles = 4.15 g/71 g/mol

Number of moles = 0.06 mol

Now we will compare the moles of AlCl₃ with Al and Cl₂.

                    Cl₂           :        AlCl₃

                    3              :          2

                   0.06         :        2/3×0.06 = 0.04

                   Al             :        AlCl₃

                     2            :          2

                   0.10         :        0.10

Number of moles of AlCl₃ produced by chlorine are less so it will be limiting reactant.

Mass of AlCl₃:Theoretical yield

Mass = number of moles ×molar mass

Mass = 0.04 mol × 133.34 g/mol

Mass = 5.33 g

4 0
1 year ago
In a fixed cylinder are 3moles of oxygen gas at 300Kelvin and 1.25atm. What is the volume of the container?
vladimir2022 [97]

Answer:

The volume of the container is 59.112 L

Explanation:

Given that,

Number of moles of Oxygen, n = 3

Temperature of the gas, T = 300 K

Pressure of the gas, P = 1.25 atm

We need to find the volume of the container. For a gas, we know that,

PV = nRT

V is volume

R is gas constant, R =  0.0821 atm-L/mol-K

So,

V=\dfrac{nRT}{P}\\\\V=\dfrac{3\ mol\times 0.0821\ L-atm/mol-K \times 300\ K}{1.25\ atm}\\\\V=59.112\ L

So, the volume of the container is 59.112 L

6 0
1 year ago
How many liters of a 0.225 M solution of KI are needed to contain 0.935 moles of KI?
Katyanochek1 [597]

Answer:

4.16L

Explanation:

From the question given, we obtained the following data:

Molarity = 0.225 M

Number of mole of KI = 0.935mole

Volume =?

Molarity = mole / Volume

Volume = mole /Molarity

Volume = 0.935/0.225

Volume = 4.16L

Therefore, 4.16L of KI is needed.

6 0
2 years ago
Imagine if during the cathode ray experiment, the size of the particles of the ray was the same as the size of the atom forming
Yakvenalex [24]

Answer:

This would support Dalton's postulates that proposed the atoms are indivisible because no small particles are involved.

Explanation:

Experiment using the gas discharge tube by J.J Thomson led to the discovery of cathode rays which are now known as electrons.

Primarily, Thomson's experiment led to the discovery of cathode rays, electrons, as subatomic particles.

If the size of the atoms observed at the cathode is the same as that of the rays,we can conclude that the particles of the rays are the simplest form of matter we can have. This would suggest that the atom is indeed the smallest indivisible particle of a matter according to Dalton.

7 0
2 years ago
Read 2 more answers
The Keq for the equilibrium below is 5.4 × 1013 at 480.0 °C. 2NO (g) + O2 (g) 2NO2 (g) What is the value of Keq at this temperat
kodGreya [7K]

<u>Answer:</u> The equilibrium constant for NO_2(g)\rightleftharpoons NO(g)+\frac{1}{2}O_2(g) equation is 1.36\times 10^{-7}

<u>Explanation:</u>

The given chemical equation follows:

2NO(g)+O_2(g)\rightleftharpoons 2NO_2(g)

The value of equilibrium constant for the above equation is K_{eq}=5.4\times 10^{13}

Calculating the equilibrium constant for the given equation:

NO_2(g)\rightleftharpoons NO(g)+\frac{1}{2}O_2(g)

The value of equilibrium constant for the above equation will be:

K'_{eq}=\frac{1}{\sqrt{K_{eq}}}\\\\K'_{eq}=\frac{1}{\sqrt{5.4\times 10^{13}}}\\\\K'_{eq}=1.36\times 10^{-7}

Hence, the equilibrium constant for NO_2(g)\rightleftharpoons NO(g)+\frac{1}{2}O_2(g) equation is 1.36\times 10^{-7}

5 0
1 year ago
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