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Nikitich [7]
3 years ago
10

8. Estimate the maximum volume percent of Methanol vapor that can exist at standard conditions. Vapor pressure = 88.5 mm Hg in a

ir.
Chemistry
1 answer:
sashaice [31]3 years ago
6 0

Explanation:

Vapor pressure is defined as the pressure exerted by vapors or gas on the surface of a liquid.

It is known that at standard condition, vapor pressure is 760 mm Hg.

And, it is given that methanol vapor pressure in air is 88.5 mm Hg.

Hence, calculate the volume percentage as follows.

                  Volume percentage = \frac{\text{given vapor pressure}}{\text{standard vapor pressure}} \times 100

                                                    = \frac{88.5}{760} \times 100

                                                    = 11.65%

Thus, we can conclude that the maximum volume percent of Methanol vapor that can exist at standard conditions is 11.65%.

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3 years ago
A mixture of bases can sometimes be the active ingredient in antacid tablets. If 0.4884 g of a mixture of Al(OH)3 and Mg(OH)2 is
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Answer:

54.7%

Explanation:

The reaction of neutralization will be:

Al(OH)₃ + Mg(OH)₂ + 5HNO₃ → Al(NO₃)₃ + Mg(NO₃)₂ + 5H₂O

The number of moles of HNO₃ added is the concentration multiplied by the volume (17.12 mL = 0.01712 L)

n = 1*0.01712 = 0.01712 mol

For the stoichiometry:

1 mol of Al(OH)₃  ---------- 5 moles of HNO₃

x ----------------------------------0.01712 mol

By a simple direct three rule:

5x = 0.01712

x = 3.424x10⁻³ mol of Al(OH)₃

For the stoichiometry, the number of moles of Mg(OH)₂ will also be 3.424x10⁻³.

The molar masses are:

Al(OH)₃: 27 g/mol of Al + 3*16 g/mol of O + 3*1 g/mol of H = 78 g/mol

Mg(OH)₂: 24 g/mol of Mg + 2*16 g/mol of O + 2*1 g/mol of H = 58 g/mol

The mass is the molar mass multiplied by the number of moles:

Al(OH)₃: 78*3.424x10⁻³ = 0.2671 g

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The mass % of Al(OH)₃ in the mixture is its mass divides by the mass of the mixture, multiplied by 100%:

(0.2671/0.4884)x100% = 54.7%

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3 years ago
In order to prepare a 0.523 m aqueous solution of potassium iodide, how many grams of potassium iodide must be added to 2.00kg o
Lerok [7]
M = amount of the solute  / mass of the <span>solvent

0.523 = x / 2.00 

x = 0.523 * 2.00

x = 1,046  moles

molar mass KI = </span><span>166.0028 g/mol
</span><span>
Mass = 1,046 * 166.0028

Mass </span>≈<span> 173.63 g

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Heart, 5 stars, and Brainiest to first right answer!
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Answer: it is c

Explanation:

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