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Crank
2 years ago
6

1. What is the mass of 2.23 x 1023 atoms of sulfur?

Chemistry
1 answer:
sertanlavr [38]2 years ago
5 0

Answer:

La masa de 2,23×10²³ átomos de azufre es 11,85 g

A partir de una comprensión detallada de la hipótesis de Avogadro, entendimos que 1 mol de cualquier sustancia contiene 6,02 × 10²³ átomos.

La afirmación anterior implica que:

6,02×10²³ átomos = 1 mol de azufre

Recordar :

1 mol de azufre = 32 g

Así, podemos decir que:

6,02×10²³ átomos = 32 g de azufre

Con la información anterior , podemos obtener la masa de 2,23×10²³ átomos de azufre. Esto se puede obtener de la siguiente manera:

6,02×10²³ átomos = 32 g de azufre

Por lo tanto,

2,23×10²³ átomos = *la imagen*

2,23×10²³ átomos = 11,85 g de azufre

Así, la masa de 2,23×10²³ átomos de azufre es 11,85 g

Explanation:

The mass of 2.23×10²³ sulfur atoms is 11.85 g

From a detailed understanding of Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02 × 10²³ atoms.

The above statement implies that:

6.02×10²³ atoms = 1 mole of sulfur

Remember :

1 mole of sulfur = 32 g

Thus, we can say that:

6.02×10²³ atoms = 32 g of sulfur

With the above information, we can obtain the mass of 2.23×10²³ sulfur atoms. This can be obtained as follows:

6.02×10²³ atoms = 32 g of sulfur

Therefore,

2.23×10²³ atoms =*the picture*

2.23×10²³ atoms = 11.85 g of sulfur

Thus, the mass of 2.23×10²³ sulfur atoms is 11.85 g

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The bond is highly polar is the scientific claim about the bond in the molecular compound HF is most likely to be true.

The term polar means that there is separation of charges, the bonds will be the more polar the greater the difference in electronegativity between the bonded atoms.

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Therefore, we can conclude that the bond is highly polar is the scientific claim about the bond in the molecular compound HF is most likely to be true.

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3 years ago
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A 1.30M solution of BaCl2 has a density of 1.230 g/ml. a) What is the mole fraction of BaCl2 in this solution?
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<u>Answer:</u> The mole fraction of barium chloride in the solution is 0.024

<u>Explanation:</u>

We are given:

Molarity of barium chloride solution = 1.30 M

This means that 1.30 moles of barium chloride is present in 1 L or 1000 mL of solution.

  • To calculate the mass of solution, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of solution = 1.230 g/mL

Volume of solution = 1000 mL

Putting values in above equation, we get:

1.230g/mL=\frac{\text{Mass of solution}}{1000mL}\\\\\text{Mass of solution}=(1.230g/mL\times 1000mL)=1230g

  • To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Moles of barium chloride = 1.30 moles

Molar mass of barium chloride = 208 g/mol

Putting values in equation 1, we get:

1.30mol=\frac{\text{Mass of barium chloride}}{208g/mol}\\\\\text{Mass of barium chloride}=(1.30mol\times 208g/mol)=270.4g

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Mass of water = 1230 - 270.4 = 959.6 g

<u>Calculating the moles of water:</u>

Given mass of water = 959.6 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

\text{Moles of water}=\frac{959.6g}{18g/mol}\\\\\text{Moles of water}=53.31mol

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\chi_A=\frac{n_A}{n_A+n_B}

  • <u>For barium chloride:</u>

\chi_{\text{(barium chloride)}}=\frac{n_{\text{(barium chloride)}}}{n_{\text{(water)}}+n_{\text{(barium chloride)}}}

\chi_{\text{(barium chloride)}}=\frac{1.30}{1.30+53.31}\\\\\chi_{\text{(barium chloride)}}=0.024

Hence, the mole fraction of barium chloride in the solution is 0.024

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Answer:

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Answer:

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Explanation:

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O_2=0.2mol/L

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NO=\frac{0.08}{0.5}

NO=0.16mol/L

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