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r-ruslan [8.4K]
4 years ago
7

1.8 x 1022 atoms of Rn to moles of Rn​

Chemistry
1 answer:
Ivenika [448]4 years ago
3 0

Answer:

1.8×10²² atom of Rn contain 0.03 moles of Rn.

Explanation:

Given data:

Number of atoms of Rn = 1.8×10²² atoms

Number of moles = ?

Solution:

we will solve this problem through 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.

For example,

1 mole of Rn = 6.022 × 10²³ atoms of Rn

1.8×10²² atoms of Rn × 1mol /  6.022 × 10²³ atoms of Rn

0.3×10⁻¹ moles of Rn

0.03 moles of Rn

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3 years ago
A hypothetical element consists of two isotopes of masses 86.95amu and 88.95amu with abundances of 35.5% and 64.5% respectively.
Art [367]

Answer: The average atomic mass of the element = 88.242amu

Explanation:

The abundance of the first isotope is =35.5%

 Atomic mass of first isotope = 68.9257

The average atomic mass of the first isotope =86.95amu X 35.5%  =86.95amu X 0.355 =30.8725 amu

The abundance of the second isotope =64.5%

Atomic mass of the second isotope =88.95amu

The average atomic mass of second isotope =88.95amu x 64.5% = 88.95amu x 0.645= 57.37275 amu

Now the average atomic mass =30.8725 +57.37275 = 88.242amu

OR using the formulae

Average atomic mass = [mass of isotope× its abundance] + [mass of isotope× its abundance] +...[ ] / 100

{(86.95amu X 35.5 )+(88.95amu x 64.5)}/100

8,824/100

=88.24amu

5 0
3 years ago
Type the correct answer in the box. a pharmacist works with a 1.75 m solution of sodium bromide (nabr) and water. the volume of
Elodia [21]

Answer: 147 mL

Explanation:

<u>Given:</u>

Molarity of the sodium bromide (NaBr) solution (M1)  = 1.75 M

Volume of the solution (V1) = 84 mL

Molarity of the diluted NaBr solution (M2) = 1 M

Using the dilution formula to solve for V2:

\begin{gathered}M_{1} V_{1}=M_{2} V_{2} \\V_{2}=(1.75 \mathrm{M} \times 84 \mathrm{~mL}) / 1 \mathrm{M} \\V_{2}=147 \mathrm{~mL}\end{gathered}

Therefore, the new volume of the solution is 147 mL

8 0
2 years ago
The pressure in an automobile tire filled with air is 245.0 kPa. If Po2 = 51.3 kPa, Pco2 = 0.10 kPa, and P-others = 2.3 kPa, wha
Kay [80]

Answer:

PN₂ = 191.3 Kpa

Explanation:

Given data:

Total pressure of tire = 245.0 Kpa

Partial pressure of PO₂ = 51.3 Kpa

Partial pressure of PCO₂  = 0.10 Kpa

Partial pressure of others =  2.3 Kpa

Partial pressure of PN₂ = ?

Solution:

According to Dalton law of partial pressure,

The total pressure inside container is equal to the sum of partial pressures of individual gases present in container.

Mathematical expression:

P(total) = P₁ + P₂ + P₃+ ............+Pₙ

Now we will solve this  problem by using this law.

P(total) = PO₂ + PCO₂ + P(others)+ PN₂

245 Kpa = 51.3 Kpa + 0.10 Kpa + 2.3 Kpa + PN₂

245 Kpa = 53.7 Kpa+ PN₂

PN₂ = 245 Kpa -  53.7 Kpa

PN₂ = 191.3 Kpa

6 0
4 years ago
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