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3 years ago

How many molecules are there in 0.661 mol of O2 gas?

Chemistry

1 answer:

QveST [7]3 years ago

5 0

Answer:

One mole of oxygen gas, which has the formula O2, has a mass of 32 g and contains 6.02 X 1023 molecules of oxygen but 12.04 X 1023 (2 X 6.02 X 1023) atoms, because each molecule of oxygen contains two oxygen atoms.

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Convert 3.52 × 10^22 atoms of gold to moles

IrinaVladis [17]

Answer:

5.85 x 10⁻² mol Au

Explanation:

Divide the atoms of gold with Avogadro's number to find the solution.

5 0

3 years ago

The vapor pressure of pure water at 25 °c is 23.8 torr. What is the vapor pressure (torr) of water above a solution prepared by

maxonik [38]

The vapour pressure of the solution is 23.4 torr.

Use Raoult’s Law to calculate the vapour pressure:

p₁ = χ₁p₁°

where

χ₁ = the mole fraction of the solvent

p₁ and p₁° are the vapour pressures of the solution and of the pure solvent

The formula for vapour pressure lowering Δp is

Δp = p₁° - p₁

Δp = p₁° - χ₁p₁° = p₁°(1 – χ₁) = χ₂p₁°

where χ₂ is the mole fraction of the solute.

Step 1. Calculate the mole fraction of glucose

n₂ = 18.0 g glu × (1 moL glu/180.0 g glu) = 0.1000 mol glu

n₁ = 95.0 g H_2O × (1 mol H_2O/18.02 g H_2O) = 5.272 mol H_2O

χ₂ = n₂/(n₁ + n₂) = 0.1000/(0.1000 + 5.272) = 0.1000/5.372 = 0.018 62

Step 2. Calculate the vapour pressure lowering

Δp = χ₂p₁° = 0.018 62 × 23.8 torr = 0.4430 torr

Step 3. Calculate the vapour pressure

p₁ = p₁° - Δp = 23.8 torr – 0.4430 torr = 23.4 torr

3 0

3 years ago

a student determined that 8.2 milligrams of oxygen is dissolved in a 1000 gram sample of water at 15 degrees Celsius and 1 atm w

melamori03 [73]

Answer: We do not know. We have not been given the solubility of oxygen in water at a given temperature nor have we been given the Henry's laws constant. We also do not know whether you mean 1 atmosphere of air, or 1 atmosphere of oxygen.

8 0

3 years ago

Ne4ueva [31]

Answer:

32,127.02 grams of hydrogen hexafluorosilicate will contain this 25,434 grams of F.

Explanation:

Volume of cylindrical reservoir = V

Radius of the cylindrical reservoir = r = d/2

d = diameter of the cylindrical reservoir = d = $4.50\times 10^1 m=45 m$

r = d/2 = 22.5 m

Depth of the reservoir = h = 10.0 m

$V=\pi r^2 h$

$=3.14\times (22.5 m)^2\times 10.0 m=15,896.25 m^3=15,896,250 L$

$1 m^3=1000 l$

Volume of water cylindrical reservoir : V

Density of water,d = 1 kg/L

Mass of water cylindrical reservoir = m

$m=d\times V=1 kg/L\times 15,896,250 L=15,896,250 kg$

1.6 kilogram of fluorine per million kilograms of water. (Given)

Concentration of fluorine in water = 1.6 kg/ 1000,000 kg of water

In 1000,000 kg of water = 1.6 kg of fluorine

Then $15,896,250 kg$ of water have x mass of fluorine:

$\frac{x}{15,896,250 kg\text{kg of water}}=\frac{1.6 kg}{1000,000 \text{kg of water}}$

$x=\frac{1.6 kg}{1000,000}\times 15,896,250 kg=25.434 kg$

15,896,250 kg water of contains mass 25.434 kg of fluorine.

25.434 kg = 25434 g

25,434 grams of fluorine should be added to give 1.6 ppm.

Percentage of fluorine in hydrogen hexafluorosilicate :

Molar mass hydrogen hexafluorosilicate = 144 g/mol

$F\%=\frac{6\times 19 g/mol}{144 g/mol}\times 100=79.16\%$

Total mass of hydrogen hexafluorosilicate = m'

$79.16\%=\frac{25,434 g}{m'}\times 100$

m' = 32,127.02 g

32,127.02 grams of hydrogen hexafluorosilicate will contain this 25,434 grams of F.

5 0

3 years ago

How did mendel use his training in mathematics to aid his experiments in life science?

Fittoniya [83]

Mendel use his training in mathematics to aid his experiments in life science by using it for the offspring sexual reproduction. He mainly studied pea plants because they had distinguished characteristics and they were quick to grow. Mendel would create hybrids from the plants.

3 0

3 years ago