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Flauer [41]
2 years ago
11

How many grams are in 1.23 x 1020 atoms of arsenic?

Chemistry
1 answer:
Alina [70]2 years ago
6 0

Given the data from the question, the mass of arsenic that contains 1.23×10²⁰ atoms is 0.0153 g

<h3>Avogadro's hypothesis </h3>

6.02×10²³ atoms = 1 mole of arsenic

But

1 mole of arsenic = 75 g

Thus, we can say that:

6.02×10²³ atoms = 75 g of arsenic

<h3>How to determine the mass that contains 1.23×10²⁰ atoms</h3>

6.02×10²³ atoms = 75 g of arsenic

Therefore,

1.23×10²⁰ atoms = (1.23×10²⁰ × 75) / 6.02×10²³ atoms)

1.23×10²⁰ atoms = 0.0153 g of arsenic

Thus, 1.23×10²⁰ atoms is present in 0.0153 g of arsenic

Learn more about Avogadro's number:

brainly.com/question/26141731

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c

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How many miles of lead are equal to 9.51 x 10^3 g Pb?
Lemur [1.5K]
Molar mass Pb = 207.2 g/mol

1 mole Pb ------------- 207.2
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hope this helps!

7 0
3 years ago
A glass flask whose volume is 1000 cm3 at a temperature of 0.800 C is completely filled with mercury at the same temperature. Wh
ira [324]

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Explanation:

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The final volume of mercury considering its volume expansion quotient = 1000 + 1000*(1.8*10^-4 *52) = 1000 + 9.36 = 1009.36 cm^3

Considering the glass as a non expanding substance, the complete excess volume of 9.36 cm3 of mercury should have overflown the container, but due to the expansion of glass, the capacity of mercury containment increases and so a lesser amount of mercury flows out.

The amount of mercury that actually flowed out = 8.50 cm3.

So, the expansion of the glass container = 9.36-8.50 = 0.86 cm3.

Using the formula for coefficient of expansion,

coefficient of volume expansion of glass = 0.86/(1000 * 52) = 0.00001654 per degree.

5 0
3 years ago
what mass of carbon dioxide gas would be produced if 10g of calcium carbonate reacted with an excess of hydrochloric acid?
LuckyWell [14K]
Answer is: 4,4 grams <span>of carbon dioxide gas would be produced.
</span>Chemical reaction: CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O.
m(CaCO₃) = 10 g.
n(CaCO₃) = 10 g ÷ 100 g/mol.
n(CaCO₃) = 0,1 mol.
From chemical reaction: n(CaCO₃) : n(CO₂) = 1 : 1.
n(CO₂) = 0,1 mol.
m(CO₂) = n(CO₂) · M(CO₂).
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m(CO₂) = 4,4 g.
7 0
3 years ago
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