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Ksju [112]
2 years ago
14

Assuming the partial pressure of oxygen in air (0.20 atm) and nitrogen in air (0.80 atm). Calculate the mole fractions of oxygen

and nitrogen in water at 298 K. FOR THIS QUESTION report the mole fraction of OXYGEN
Chemistry
1 answer:
Nuetrik [128]2 years ago
6 0

Answer:

oxygen = 4.7 * 10^-6

Nitrogen = = 9.7 * 10^6

Explanation:

partial pressure of oxygen = 0.20 atm

partial pressure of Nitrogen = 0.80 atm

<u>calculate the mole fractions of oxygen and Nitrogen in water </u>

Temp = 298k

applying henry's law

molar conc of oxygen in water ( Coxygen )

= Kp = 1.3 * 10^-3 Mol/L.atm * 0.20 atm = 2.6 * 10^-4 Mol

molar conc of Nitrogen in water ( Cnitrogen )

= Kp = 6.8 * 10^-4 Mol/L.atm * 0.80 atm = 5.4 * 10^-4

next Given that the number of moles in 1 liter of water = 55.5 mol  

therefore the mole fraction of oxygen

= 2.6 * 10^-4 / 55.5

= 4.7 * 10^-6

mole fraction of Nitrogen

= 5.4 * 10^-4  / 55.5

= 9.7 * 10^6

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Answer:the 5g would be more dense due to the mass and structure

Of it causing the atoms to expand as heat gets added

Explanation:

8 0
3 years ago
A gold coin contains 3.47 × 10^23<br> gold atoms.<br> What is the mass of the coin in grams?
VashaNatasha [74]

Answer:

Mass = 114.26 g

Explanation:

Given data:

Number of gold atoms = 3.47×10²³ atoms

Mass in gram = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.  The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ atoms

3.47×10²³ atoms × 1 mol  /6.022 × 10²³ atoms

0.58 mol

Mass of gold:

Mass = number of moles × molar mass

Mass = 0.58 mol × 197 g/mol

Mass = 114.26 g

4 0
3 years ago
2. Calculate the mass of 3.47x1023 gold atoms.
lapo4ka [179]

3.47 x 10^{23} atoms of gold have mass of 113.44 grams.

Explanation:

Data given:

number of atoms of gold = 3.47 x 10^{23}

mass of the gold in given number of atoms = ?

atomic mass of gold =196.96 grams/mole

Avagadro's number = 6.022 X 10^{23}

from the relation,

1 mole of element contains 6.022 x 10^{23} atoms.

so no of moles of gold given = \frac{3.47 X 10^{23}  }{6.022 X 10^{23} }

0.57 moles of gold.

from the relation:

number of moles = \frac{mass}{atomic mass of 1 mole}

rearranging the equation,

mass = number of moles x atomic mass

mass = 0.57 x 196.96

mass = 113.44 grams

thus, 3.47 x 10^{23} atoms of gold have mass of 113.44 grams

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