For every liter of sea water that evaporates, 3.7 g of magnesium hydroxide are produced. How many liters of sea water must evaporate to produce 5.00 moles of magnesium hydroxide?
Answers
The liters of sea water must evaporate to produce 5.00 moles of magnesium hydroxide - 78.8 litres.
A mole is a special unit of quantity of a chemical species that contains exactly Avogadro's number of particles.
- The mass of one mole of an element or a compound, called the molar mass, can be found from a periodic table.
- The unit of molar mass is g/mol.
=> the molar mass of Mg(OH)₂
Mg + 2O + 2H
= (24.31) + (2×16) + (2×1.01)
= 58.33 gmol⁻¹.
then,
=> the number of moles:
n = m/M
=
= 0.0634 mol
=> the amount of water in liters needed to evapourate to produce 5.00 moles:
=>
=> 0.0634x = 5×1
=>
=> x = 78.8 litres.
Thus, the liters of sea water must evaporate to produce 5.00 moles of magnesium hydroxide - 78.8 litres.
learn more:
First we find the molar mass of Mg(OH)₂
Mg + 2O + 2H = 24.31 + 2×16 + 2×1.01 = 58.33 gmol⁻¹.
Then find the number of moles: n = m/M = 3.7 / 58.33 = 0.0634 mol
So now we know that 0.0634 mol are produced per litre of sea water.
Let the amount of water in litres needed to evapourate to produce 5.00 moles be x.
We can express this as a ratio.
0.0634 mol / 1 l = 5 mol / x l.
Solve for x by cross-multiplication:
0.0634x = 5×1
x = 5/0.0634 = 78.8 litres. (rounded to three significant figures)
Hope this helps!
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Final answer:
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Explanation:
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Based on this, the formulae of the ions present in this salt could be:
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So, the salt might consist of H+ cations and a combination of N3- and O2- anions, depending on the specific compound.
Answers
The partial pressure of each component of the gas are:
1. The partial pressure of CH₄ is 276.48 KPa
2. The partial pressure of C₂H₆ is 27.34 KPa
3. The partial pressure of C₃H₈ is 3.38 KPa
We'll begin by calculating the mole fraction of each gas.
For CH₄:
Percentage of CH₄ = 90%
Total = 100%
Mole fraction CH₄ =?
Mole fraction = mole / total
Mole fraction CH₄ = 90 / 100
Mole fraction CH₄ = 0.9
For C₂H₆:
Percentage of C₂H₆ = 8.9%
Total = 100%
Mole fraction C₂H₆ =?
Mole fraction = mole / total
Mole fraction C₂H₆ = 8.9 / 100
Mole fraction C₂H₆ = 0.089
For C₃H₈:
Percentage of C₃H₈ = 1.1%
Total = 100%
Mole fraction C₃H₈ =?
Mole fraction = mole / total
Mole fraction C₃H₈ = 1.1 / 100
Mole fraction C₃H₈ = 0.011
Finally, we shall determine the partial pressure of each gas. This can be obtained as follow:
1. Determination of the partial pressure of CH₄
Mole fraction CH₄ = 0.9
Total pressure = 307.2 KPa
Partial pressure of CH₄ =?
Partial pressure = mole fraction × Total pressure
Partial pressure of CH₄ = 0.9 × 307.2
Partial pressure of CH₄ = 276.48 KPa
2. Determination of the partial pressure of C₂H₆
Mole fraction C₂H₆ = 0.089
Total pressure = 307.2 KPa
Partial pressure of C₂H₆ =?
Partial pressure = mole fraction × Total pressure
Partial pressure of C₂H₆ = 0.089 × 307.2
Partial pressure of C₂H₆ = 27.34 KPa
3. Determination of the partial pressure of C₃H₈
Mole fraction C₃H₈ = 0.011
Total pressure = 307.2 KPa
Partial pressure of C₃H₈ =?
Partial pressure = mole fraction × Total pressure
Partial pressure of C₃H₈ = 0.011 × 307.2
Partial pressure of C₃H₈ = 3.38 KPa
Learn more: brainly.com/question/15754440
Answer:
276.48 atm → CH₄
27.3 atm → C₂H₆
3.38 atm → C₃H₈
Explanation:
Percentages of each gas, are the mole fraction
0.9 CH₄
0.089 C₂H₆
0.011 C₃H₈
Mole fraction = Partial pressure each gas/ Total pressure
0.9 = Partial pressure CH₄ / 307.2 kPa
307.2 kPa . 0.9 = 276.48 atm
0.089 = Partial pressure C₂H₆ / 307.2 kPa
307.2 kPa . 0.089 = 27.3 atm
0.011 = Partial pressure C₃H₈ / 307.2 kPa
307.2 kPa . 0.011 = 3.38 atm
Answers
b. boiling-point elevation
c. vapor-pressure lowering
d. freezing-point depression
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Answer:
its D
Explanation:
i just got it correct on usatestprep