Which SI units can be used to describe an object’s volume?

telo118 [61]

The correct answer is letter (A) Acetylene. Acetylene is the most simplest form of alkyne and at the same time a hydrocarbon. It is unsaturated because of the presence of only two carbon atoms that are bonded together in a triple bond. I<span>n its pure, it is unstable and thus, it is usually held and handled as a solution.</span>

4 0

2 years ago

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A plastic bottle is closed outdoors on a cold day when the temperature is −15.0°C and is later brought inside where the temperat

Lana71 [14]

Answer:

Pressure = 1.14 atm

Explanation:

Hello,

This question requires us to calculate the final pressure of the bottle after thermal equilibrium.

This is a direct application of pressure law which states that in a fixed mass of gas, the pressure of a given gas is directly proportional to its temperature, provided that volume remains constant.

Mathematically, what this implies is

P = kT k = P / T

P1 / T1 = P2 / T2 = P3 / T3 =........= Pn / Tn

P1 / T1 = P2 / T2

P1 = 1.0atm

T1 = -15°C = (-15 + 273.15)K = 258.15K

P2 = ?

T2 = 21.5°C = (21.5 + 273.15)K = 294.65K

P1 / T1 = P2 / T2

P2 = (P1 × T2) / T1

P2 = (1.0 × 294.65) / 258.15

P2 = 1.14atm

The pressure of the gas after attaining equilibrium is 1.14atm

3 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

2 years ago

Find out no. of moles present is 72 gm of water?

Alona [7]

72g H2O x 1 mol H2O/18.02g H2O = 3.99 mol H2O

8 0

2 years ago

What is the base dissociation constant for a weak base at equilibrium, B + H2O <===> BH+ + OH–?

podryga [215]

Depends on what the base is. You would reference the base dissociation chart for that value.

3 0

2 years ago

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