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beks73 [17]
3 years ago
15

What is the mass of 1.33 moles of Ba3(PO4)2?

Chemistry
1 answer:
Serga [27]3 years ago
8 0

Answer:

801 g

Explanation:

From the question given above, the following data were obtained:

Number of mole of Ba₃(PO₄)₂ = 1.33 moles

Mass of Ba₃(PO₄)₂ =?

Next, we shall determine the molar mass of Ba₃(PO₄)₂. This can be obtained as follow:

Molar mass of Ba₃(PO₄)₂ = (137.3×3) + 2[31 + (4×16)]

= 411.9 + 2[31 + 64]

= 411.9 + 2[95]

= 411.9 + 190

Molar mass of Ba₃(PO₄)₂ = 601.9 g/mol

Finally, we shall determine the mass of Ba₃(PO₄)₂. This can be obtained as follow:

Number of mole of Ba₃(PO₄)₂ = 1.33 moles

Molar mass of Ba₃(PO₄)₂ = 601.9 g/mol

Mass of Ba₃(PO₄)₂ =?

Mole = mass /Molar mass

1.33 = Mass of Ba₃(PO₄)₂ / 601.9

Cross multiply

Mass of Ba₃(PO₄)₂ = 1.33 × 601.9

Mass of Ba₃(PO₄)₂ = 801 g

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Lilit [14]
We can use the ideal gas law equation to find the pressure 
PV = nRTwhere 
P - pressure 
V - volume  - 2.6 x 10⁻³ m³ 
n - number of moles - 0.44 mol
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 25 °C + 273 = 298 K
substituting the values into the equation,
P x 2.6 x 10⁻³ m³  = 0.44 mol x 8.314 Jmol⁻¹K⁻¹ x 298 K
P = 419 281.41 Pa
101 325 Pa is equivalent to 1 atm 
Therefore 419 281.41 Pa - 1/ 101 325 x 419 281.41 = 4.13 atm
Pressure is 4.13 atm
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Answer:

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