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3 years ago

How many hydrogen atoms are in 5.80 mol of ammonium sulfide

Chemistry

1 answer:

IRINA_888 [86]3 years ago

5 0

Moles of ammonium sulfide = 5.80 mol

The formula of ammonium sulfide is (NH₄)₂S

So each molecule of ammonium sulfide has (4 x 2) or 8 atoms of H

One mole of ammonium sulfide has 8 moles of H

5.80 mol of ammonium sulfide has (8 x 5.8) or 46.4 moles of H

As per the definition of Avogadro's number, 1 mole = 6.022 x 10²³ atoms

46.4 moles of H x (6.022 x 10²³ atoms/ 1 mole of H)

= 2.8 x 10²⁵ H atoms

Therefore, 2.8 x 10²⁵ H atoms are in 5.80 mol of ammonium sulfide.

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What are some of the forms of energy you used today? How did you use them?

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Answer:

Heat or Thermal energy, Solar Energy, Chemical energy, electrical energy, mechanical energy

Explanation:

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3 years ago

Why is there a difference in the conductivity of pure solid NaC1 and of the 1.0 M NaC1 soulution

Basile [38]

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3 years ago

Convert from moles to particles 6.02x1023 particles 1 mole Question 2. 2.7 moles of lithium Question 3. 1.8 moles of sodium chlo

spayn [35]

Considering the definition of Avogadro's number:

the number of molecules of lithium is 1.62621×10²⁴ molecules.
the number of molecules of sodium chloride is 1.08414×10²⁴ molecules.

<h3>Definition of Avogadro's number</h3>

Avogadro's Number or Avogadro's Constant is called the number of particles that make up a substance (usually atoms or molecules) and that can be found in the amount of one mole of said substance. Its value is 6.023×10²³ particles per mole. Avogadro's number applies to any substance.

<h3>Amount of molecules in this case</h3>

You can apply the following rule of three, considering the Avogadro's number: If 1 mole of lithium contains 6.023×10²³ molecules, 2.7 moles of lithium contains how many molecules?

amount of molecules of lithium= (6.023×10²³ molecules × 2.7 moles)÷1 mole

<u><em>amount of molecules of lithium=1.62621×10²⁴ molecules</em></u>

Finally, the number of molecules of lithium is 1.62621×10²⁴ molecules.

On the other hand you can apply the following rule of three, considering the Avogadro's number: If 1 mole of sodium chloride contains 6.023×10²³ molecules, 1.8 moles of sodium chloride contains how many molecules?

amount of molecules of sodium chloride= (6.023×10²³ molecules × 1.8 moles)÷1 mole

<u><em>amount of molecules of sodium chloride=1.08414 ×10²⁴ molecules</em></u>

Finally, the number of molecules of sodium chloride is 1.08414×10²⁴ molecules.

Learn more about Avogadro's Number:

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brainly.com/question/1445383

brainly.com/question/1528951

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2 years ago

At 35.0°c and 3.00 atm pressure, a gas has a volume of 1.40 l. what pressure does the gas have at 0.00°c and a volume of 0.950 l

Leona [35]

Answer : The pressure of gas will be, 3.918 atm and the combined gas law is used for this problem.

Solution :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 3 atm

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = 1.40 L

$V_2$ = final volume of gas = 0.950 L

$T_1$ = initial temperature of gas = $35^oC=273+35=308K$

$T_2$ = final temperature of gas = $0^oC=273+0=273K$

Now put all the given values in the above equation, we get the final pressure of gas.

$\frac{3atm\times 1.40L}{308K}=\frac{P_2\times 0.950L}{273K}$

$P_2=3.918atm$

Therefore, the pressure of gas will be, 3.918 atm and the combined gas law is used for this problem.

4 0

3 years ago

Read 2 more answers

A solution contains 0.60 mg/ml mn2+. what minimum mass of kio4 must me added to 5.00 ml of the solution in order to completely o

azamat

The given solution of Mn²⁺ is 0.60 mg/mL.
Hence mass of Mn²⁺ in 5 mL of solution = 0.60 mg/mL x 5 mL = 3 mg

Molar mass of Mn = 54.9 g/mol
Hence, moles of Mn²⁺ = 3 x 10⁻³ g / 54.9 g/mol = 5.46 x 10⁻⁵ mol

The balanced equation for the reaction is,
2Mn²⁺ + 5KIO₄ + 3H₂O → 2MnO₄⁻ + 5KIO₃ + 6H⁺

The stoichiometric ratio between Mn²⁺ and KIO₄ is 2 : 5

Hence, moles of KIO₄ reacted = 5.46 x 10⁻⁵ mol x (5 / 2)
= 13.65 x 10⁻⁵ mol
Molar mass of KIO₄ = 230 g/mol

Hence needed mass of KIO₄ = 13.65 x 10⁻⁵ mol x 230 g/mol
= 0.031395 g
= 31.395 mg
≈ 31.4 mg

5 0

3 years ago