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zlopas [31]
3 years ago
9

5. A salvage operator recovered coins believed to be gold. A sample weighed 129.6 g and had a volume of 15.3 cm3. Were the coins

gold (d= 19.3 g/cm3) or just yellow brass (d= 8.47 b/cm3)?
Chemistry
1 answer:
Nat2105 [25]3 years ago
8 0

Answer:

Therefore the coins were of yellow brass.

Explanation:

A sample weighted 129.6 g and had a volume of 15.3 cm³.

Density =\frac{mass }{volume}

Here mass= 129.6 g and volume =15.3 cm³.

Density =\frac{129.6}{15.3} g/cm³

             =8.47 g/cm³

Given that the density of gold = 19.3 g/cm³

and density of yellow brass = 8.47 g/cm³

Since the density of the sample is equal to the density of yellow brass.

Therefore the coins were of yellow brass.

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<h3>What are Isomers ?</h3>

Isomers are compounds which have same empirical formula , same number of atoms are present but they have difference in arrangement if the atoms.

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There are four cis, trans isomers for 2-isopropyl-5-methylcyclohexanol

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4. Suppose 8.00 g of CH4 is allowed to burn in the presence of 16.00 g of oxygen. CH4(g)+2O2(g)--&gt;CO2(g)+2H2O(g)
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<h3>Answer:</h3>

No masses of CH₄ and O₂ remained after the reaction, while 22.005 g of CO₂ and 18.02 g of H₂O remained

<h3>Explanation:</h3>

The combustion of methane is given by the reaction;

CH₄(g)+2O₂(g) → CO₂(g)+2H₂O(g)

We are given, 8 g of CH₄ and 16.00 g of O₂

Required to determine the mass of CH₄, O₂, CO₂ and H₂O that remained after the complete reaction.

<h3>Step 1: Moles of CH₄ and O₂ in the mass given </h3>

Moles = mass ÷ molar mass

Molar mass of CH₄ = 16.04 g/mol

Moles of CH₄ = 8.00 g ÷ 16.04 g/mol

                      = 0.498 moles

                      = 0.5 moles

Molar mass of O₂ = 16.0 g/mol

Moles of O₂ = 16.00 g ÷ 16.00 g/mol

                    = 1 mole

From the reaction, 1 mole of CH₄ reacts with 2 moles of O₂

CH₄ is the limiting reactant since it is way less than the amount of O₂

Therefore, 0.5 moles of CH₄ will react with 1 mole of oxygen.

This means there will be no amount of O₂ and CH₄ that remains.

<h3>Step 2: Moles of CO₂ and H₂O that were produced.</h3>

From the reaction 1 mole of CH₄ reacts with 2 moles of O₂ to produce 1 mole of CO₂ and 2 moles of H₂O.

Therefore,

In our case, 0.5 moles of CH₄ will react with 1 mole of O₂ to produce 0.5 moles of CO₂ and 1 mole of H₂O.

<h3>Step 3: Mass of CO₂ and H₂O produced </h3>

Mass = Moles × Molar mass

Molar mass of CO₂ = 44.01 g/mol

Mass of CO₂ = 0.5 mol × 44.01 g/mol

                      = 22.005 g

Molar mass of H₂O = 18.02 g/mol

Moles of H₂O = 1 mole × 18.02 g/mol

                       = 18.02 g

Therefore, no masses of CH₄ and O₂ remained after the reaction, 22.005 g of CO₂ and 18.02 g of H₂O remained

3 0
4 years ago
Read 2 more answers
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