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3 years ago

How many hydrogen atoms are in 4.40 mol of ammonium sulfide

2 answers:

katrin2010 [14]3 years ago

5 0

The number of hydrogen atoms that are in 4.40 mol of ammonium sulfide is 2.12 x10^25 atoms

calculation
find the number of moles of Hydrogen in ammonium sulfide (NH4)2S

that is 4.40 x number of hydrogen atoms in (NH4)2S ( 4x2= 8 atoms)

moles is therefore= 4.40 x8= 35.2 moles

by use of Avogadro's law constant

that is 1mole = 6.02 x10^23 atoms
35.2 moles=?

by cross multiplication
{35.2 moles x 6.02 x10^23} /1 mole = 2.12 x10^25 atoms

Ne4ueva [31]3 years ago

4 0

The molecular formula for ammonium sulfide would be (NH4)2S. That means in one molecule of ammonium sulfide there will be 1 N + 4 H + 2 S molecules.
Then the amount of hydrogen atoms in 4.4 moles of ammonium sulfide should be: 4.4 moles * 4/1= 17.6 moles
If you are asked for atom count then the amount should be: 17.6 moles * 6.022x10^23 atoms/mol = 105.9872 * 10^23= 1.06 * 10^25

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Answer:

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Explanation:

The absorbance of a copper sulfate solution can be calculated using Beer-Lambert Law:

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3 years ago

How is the wavelength of an electromagnetic wave related to the energy of the wave?

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3 years ago

Read 2 more answers

1. A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0%

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Answer:

The mass of PbI2 will be 18.2 grams

Explanation:

Step 1: Data given

Volume solution = 99.8 mL = 0.0998 L

mass % KI = 12.0 %

Density = 1.093 g/mL

Volume of the other solution = 96.7 mL = 0.967 L

mass % of Pb(NO3)2 = 14.0 %

Density = 1.134 g/mL

Step 2: The balanced equation

Pb(NO3)2(aq) + 2 KI(aq) ⇆ PbI2(s) + 2 KNO3(aq)

Step 3: Calculate mass

Mass = density * volume

Mass KI solution = 1.093 g/mL * 99.8 mL

Mass KI solution = 109.08 grams

Mass KI solution = 109.08 grams *0.12 = 13.09 grams

Mass of Pb(NO3)2 solution = 1.134 g/mL * 96.7 mL

Mass of Pb(NO3)2 solution = 109.66 grams

Mass of Pb(NO3)2 solution = 109.66 grams * 0.14 = 15.35 grams

Step 4: Calculate moles

Moles = mass / molar mass

Moles KI = 13.09 grams / 166.0 g/mol

Moles KI = 0.0789 moles

Moles Pb(NO3)2 = 15.35 grams / 331.2 g/mol

Moles Pb(NO3)2 = 0.0463 moles

Step 5: Calculate the limiting reactant

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

Ki is the limiting reactant. It will completely be consumed ( 0.0789 moles). Pb(NO3)2 is in excess. There will react 0.0789/2 = 0.03945 moles. There will remain 0.0463 - 0.03945 = 0.00685 moles

Step 6: Calculate moles PbI2

For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3

For 0.0789 moles KI we'll have 0.0789/2 = 0.03945 moles PbI2

Step 7: Calculate mass of PbI2

Mass PbI2 = moles PbI2 * molar mass PbI2

Mass PbI2 = 0.03945 moles * 461.01 g/mol

Mass PbI2 = 18.2 grams

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