Answer:
the partial pressure of Xe is 452.4 mmHg
Explanation:
Dalton's law of partial pressures says that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of the individual gases.
The partial pressures can be calculated with the molar fraction of the gas, in this case, Xe.
Molar fraction of Xe is calculated as follows:
Then, 0.29 is the molar fraction of Xe in the mixture of gases given.
To know the parcial pressure of Xe, we have to multiply the molar fraction by the total pressure:
Partial Pressure of Xe=1560mmHg*0.29
Partial Pressure of Xe=452.4mmHg
Answer:
- <em>The unknown integer X in the formula is </em><u>5</u><em>.</em>
Explanation:
<u>1) Data:</u>
a) Mass of CuSO₄ ∙ XH₂O = 1.50 g
b) Mass of CuSO₄ = 0.96
c) X = ?
<u>2) Additional needed data:</u>
a) Molar Mass of CuSO₄ = 159,609 g/mol
b) Molar mass of H₂O = 18,01528 g/mol
<u>3) Chemical principles and formulae used:</u>
a) Law of conservation of mass
b) Molar mass = mass in grams / number of moles = m / n
<u>4) Solution:</u>
a) Law of conservation of mass:
- Mass of CuSO₄ ∙ XH₂O = mass of CuSO₄ + mass of H₂O
- 1.50g = 0.96g + mass of H₂O ⇒ mass of H₂O = 1.50g - 0.96g = 0.54g
b) Moles
- CuSO₄: n = 0.96g / 159.609 g/mol = 0.0060 mol
- H₂O: 0.54g / 18.01528 g/mol = 0.030 mol
c) Proportion:
Divide both mole amounts by the least of the two numbers, i.e. 0.0060
- CuSO₄: 0.0060 / 0.0060 = 1
- H₂O: 0.030 mol / 0.0060 = 5
Then, the ratio of CuSO₄ to H₂O is 1 : 5 and the chemical formula is:
Hence, the value of X is 5.
Answer:
0.60 mol·L⁻¹
Explanation:
Data:
LiBr: c = 0.50 mol/L; V =300 mL
RbBr: c = 0.70 mol/L; V =300 mL
1. Calculate the moles of Br⁻ in each solution
(a) LiBr
(b) RbBr
2. Calculate the molar concentration of Br⁻
(a) Moles of Br⁻
n = 0.150 mol + 0.210 mol = 0.360 mol
(b) Volume of solution
V = 300 mL + 300 mL = 600 mL = 0.600 L
(c) Molar concentration
Answer:
I would say the correct answer is <em><u>A 200.00 mol</u></em>
Explanation:
Hope this help
Have a nice day
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