All categories

2 years ago

If I have 0.275 moles of gas at a temperature of 75 K and a pressure of 1530 mmHg, what is the volume of the gas?

Chemistry

1 answer:

Maru [420]2 years ago

6 0

Answer: The volume of the gas is 0.84 Liters

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 1530 mm Hg = 2.01 atm (760mmHg=1atm)

V = Volume of gas = ?

n = number of moles = 0.275

R = gas constant = $0.0821Latm/Kmol$

T =temperature = $75K$

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

$V=\frac{0.275atm\times 0.0820 Latm/K mol\times 75K}{2.01atm}=0.84L$

Thus the volume of the gas is 0.84 Liters

You might be interested in

When doing a recrystallization what observations would prompt

Volgvan

<h2>Observation of purified crystals.</h2>

Explanation:

Recrystallization is the process in which a compound is dissolved in a hot solvent and then allowed to cool which forms the crystals.
It is a technique used for purification of solids.
Recrystallization mainly depends on the solubility of the solvent that means the solid should be soluble in the solvent when it is hot and should be insoluble in the same solvent when it is cool.
The observations while doing recrystallization is to observe the purified crystals obtained and to filter it out.

5 0

3 years ago

The following are electronegativity. What value should be where the M is?0.981.57M2.553.043.443.98No data

ipn [44]

Answer 2.04

Explanation

Electronegativity decreases down the group and increases across the period in the periodic table.

In the period two of the periodic table,we have the following values for electronegativities with respect to its elements.

Li...........0.98

Be.........1.57

B..........2.04

C.........2.55

N..........3.04

O...........3.44

F...........3.98

Ne........n.a

The value that should be where M is is 2.04

6 0

1 year ago

C3H8+3O2 = 3CO2+4H2O what is the enthalpy combustion

dezoksy [38]

Answer:

$\Delta H_{comb}=2043.85kJ/mol$

Explanation:

Hello there!

In this case, according to the given chemical reaction, it possible for us to set up the expression for the calculation of the enthalpy change as shown below:

$\Delta H_r=-\Delta H_{comb}=3\Delta _fH_{CO_2}+4\Delta _fH_{H_2O}-\Delta _fH_{C_3H_8}-3\Delta _fH_{O_2}$

Thus, given the values of the enthalpies of formation on the attached file, we obtain: $-\Delta H_{comb}=3(-393.5kJ/mol)+4(-241.8kJ/mol)-(-103.85kJ/mol)-3(0kJ/mol)\\\\-\Delta H_{comb}=-2043.85kJ/mol\\\\\Delta H_{comb}=2043.85kJ/mol$