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3 years ago

What volume is occupied by 18.4 g oxygen at 28.0°C and a pressure of 0.998 torr? (round to sig figs)

Chemistry

1 answer:

Alexxandr [17]3 years ago

8 0

Answer:

11.0 dm³

Explanation:

From the question,

Applying

PV= nRT............... Equation 1

Where P = pressure of oxygen gas, V = volume of oxygen gas, n = number of moles of oxygen, R = molar constant, T = Temperature.

make V the subeject of the equation

V = nRT/P............. Equation 2

But,

Number of mole (n) = Mass of oxygen(m)/Molar mass of oxygen(m')

n = m/m'....................... Equation 3

Substitute equation 3 into equation 2

V = mRT/Pm'............. Equation 4

Given: T = 28°C = (28+273) = 301 K, P = 0.998 torr = (0.998×0.00131579) = 1.3132 atm, m = 18.4 g

Constant: R = 0.082 atm.dm³/K.mol, m' = 32 g/mol.

Substitute these values into equation 4

V = (301×18.4×0.082)/(32×1.3132)

V = 454.1488/42.0224

V = 10.81 dm³

V = 11.0 dm³

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The value of ΔG° at 181.0 °C for the formation of calcium chloride from calcium metal and chlorine gas is ________ kJ/mol. At 25

aleksandr82 [10.1K]

Answer:

$\Delta G^{0}$ at 181.0 $^{0}\textrm{C}$ is -723.3 kJ/mol.

Explanation:

We know, $\Delta G^{0}=\Delta H^{0}-T\Delta S^{0}$

where, T is temperature in kelvin.

Let's assume $\Delta H^{0}$ and $\Delta S^{0}$ does not change in the temperature range 25.0 $^{0}\textrm{C}$ - 181.0 $^{0}\textrm{C}$ .

$181.0^{0}\textrm{C}$ = (273+181.0) K = 454.0 K

Hence, at 181.0 $^{0}\textrm{C}$ , $\Delta G^{0}=(-795.8kJ/mol)-[(454.0 K)\times (-159.8\times 10^{-3}kJ/K.mol)]$

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3 years ago

what is the molar mass of a gaseous flouride of sulfur containing 70.4% F and having a density of approximately 4.5g/L at 20 deg

zlopas [31]

Answer:

The molar mass is 180.2 g/mol

Explanation:

<u>Step 1:</u> Data given

% of F = 70.4 %

Density = 4.5 g/L

Temperature = 20 °C

Pressure = 1 atm

<u>Step 2:</u> Calculate the number of moles

PV = nRT

⇒ with P = the pressure = 1.00 atm

⇒ with V = the volume = Assume this is 1L

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

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

⇒ T = the temperature : 20°C = 293 Kelvin

1 atm*1L= n(0.08206 L-atm/mol-K)*(293 K)

n = 0.04159 moles

<u>Step 3</u>: Calculate molar mass

Molar mass = Mass / moles

4.5 grams / 0.04159 moles = 108.2 g/mol

<u>Step 4:</u> Calculate moles of F

Moles = Mass / molar mass

Moles F = 70.4 g / 19 g/mol

Moles F = 3.70 moles

Moles S = 29.6g / 32.07 g/mol

Moles S = 0.923 moles S

<u>Step 5:</u> Divide by the smallest amount of moles

F = 3.70 / 0.923 = 4

S = 0.923 / 0.923 = 1

The empirical formula is SF4

The molar mass of SF4 = 32.07 + 4*19 = 108.07 g/mol

This means the empirical formula is the same as the molecular formula SF4

The molar mass is 180.2 g/mol

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