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Luba_88 [7]
3 years ago
12

< Progress: 19/21 groups Due Jan 15 at 11:55 PM Use the References to access important values if needed for this question. Th

e compound cobalt (II) sulfate forms a hydrate with seven water molecules per formula unit. What are the name and formula of the hydrate
Chemistry
1 answer:
Ierofanga [76]3 years ago
6 0

Answer:

CoSO4.7H2O

Cobalt II tetraoxosulphate VI heptahydrate

Explanation:

According to IUPAC nomenclature, compounds are named systematically.

We were told that there are seven water molecules in each formula unit Hence the correct formula of the compound is CoSO4.7H2O

According to IUPAC system, the metal is first named, followed by its oxidation state in Roman numerals. Then the name of the anion is mentioned as well as the number of water molecules. We have to take into account the correct prefix signifying the number of molecules of water of crystallization.

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What is the composition, in atom percent, of an alloy that contains a) 44.5 lbm of silver, b) 84.7 lbm of gold, and c) 7.3 lbm o
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<u>Answer:</u> The atom percent of silver, gold and copper in the alloy is 43.08 %, 44.93 % and 11.99 % respectively

<u>Explanation:</u>

To convert the given masses into grams, we use the conversion factor:

1 lb = 453.6 grams

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

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

To calculate the atom percent of substance in sample, we use the equation:

\%\text{ composition of substance}=\frac{\text{Moles of substance}\times N_A}{\text{Total moles of sample}\times N_A}\times 100     ......(2)

where,

N_A = Avogadro's number

  • <u>Moles of Silver:</u>

Mass of silver = 44.5 lb = 20185.2 grams

We know that:

Molar mass of silver = 107.87 g/mol

Putting values in equation 1, we get:

\text{Moles of silver}=\frac{20185.2g}{107.87g/mol}=187.12mol

  • <u>Moles of Gold:</u>

Mass of gold = 84.7 lb = 38419.9 grams

We know that:

Molar mass of gold = 196.97 g/mol

Putting values in equation 1, we get:

\text{Moles of gold}=\frac{38419.9g}{196.97g/mol}=195.05mol

  • <u>Moles of Copper:</u>

Mass of copper = 7.3 lb = 3311.3 grams

We know that:

Molar mass of copper = 63.55 g/mol

Putting values in equation 1, we get:

\text{Moles of copper}=\frac{3311.3g}{63.55g/mol}=52.10mol

Total moles of the sample =

  • <u>For Silver:</u>

Moles of silver = 187.12 moles

Total moles = [187.12 + 195.05 + 52.10] = 434.27 moles

Putting values in equation 2, we get:

\%\text{ composition of silver}=\frac{187.12\times N_A}{434.27\times N_A}\times 100\\\\\%\text{ composition of silver}=43.08\%

  • <u>For Gold:</u>

Moles of gold = 195.05 moles

Total moles = [187.12 + 195.05 + 52.10] = 434.27 moles

Putting values in equation 2, we get:

\%\text{ composition of gold}=\frac{195.05\times N_A}{434.27\times N_A}\times 100\\\\\%\text{ composition of gold}=44.93\%

  • <u>For Copper:</u>

Moles of copper = 52.10 moles

Total moles = [187.12 + 195.05 + 52.10] = 434.27 moles

Putting values in equation 2, we get:

\%\text{ composition of copper}=\frac{52.10\times N_A}{434.27\times N_A}\times 100\\\\\%\text{ composition of copper}=11.99\%

Hence, the atom percent of silver, gold and copper in the alloy is 43.08 %, 44.93 % and 11.99 % respectively

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3 years ago
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How does the law of conservation of mass apply  to this reaction: C2H4 + O2 → H2O + CO2?

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