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Leokris [45]
3 years ago
8

Silver tarnishes as silver metal reacts with hydrogen sulfide, H2S, in the air. In this reaction, dark silver sulfide, Au2S, cov

ers the surface of silver. When silver is polished, this coating of silver sulfide can be removed from the surface. This makes the silver shiny again. Enter the coefficients that balance the tarnishing reaction equation. (Type 1 for no coefficient.)Ag(s) + H2S(g) → Ag2S(s) + H2(g)
Chemistry
2 answers:
Rudik [331]3 years ago
8 0
2Ag(s) + H₂S(g) → Ag₂S(s) + H₂(g)
DaniilM [7]3 years ago
6 0

Balanced the tarnishing equation:

<u>2</u>Ag(s) + <u>1</u>H₂S(g) → <u>1</u>Ag₂S(s) + <u>1</u>H₂(g).

According to principle of mass conservation, number of atoms must be equal on both side of balanced chemical reaction.

There are two silver atoms, two hydrogen atoms and one sulfur atom on both side of balanced chemical reaction.

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A. a wide variety of hunting techniques

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3 years ago
Which is greater? 1,000ml or 1,000 cm3
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Enter your answer in the box provided. How many grams of helium must be added to a balloon containing 6.24 g helium gas to doubl
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Answer : The mass of helium gas added must be 12.48 grams.

Explanation : Given,

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First we have to calculate the moles of helium gas.

\text{Moles of }He=\frac{\text{Mass of }He}{\text{Molar mass of }He}=\frac{6.24g}{4g/mole}=1.56moles

Now we have to calculate the moles of helium gas at doubled volume.

According to the Avogadro's law, the volume of gas is directly proportional to the number of moles of gas at same pressure and temperature. That means,

V\propto n

or,

\frac{V_1}{V_2}=\frac{n_1}{n_2}

where,

V_1 = initial volume of gas  = V

V_2 = final volume of gas = 2V

n_1 = initial moles of gas  = 1.56 mole

n_2 = final moles of gas  = ?

Now we put all the given values in this formula, we get

\frac{V}{2V}=\frac{1.56mole}{n_2}

n_2=3.12mole

Now we have to calculate the mass of helium gas at doubled volume.

\text{Mass of }He=\text{Moles of }He\times \text{Molar mass of }He

\text{Mass of }He=3.12mole\times 4g/mole=12.48g

Therefore, the mass of helium gas added must be 12.48 grams.

4 0
3 years ago
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