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3 years ago

Why might the amount of copper produced be less than 100% of the expected amount? Check all possible reasons.

Chemistry

2 answers:

kkurt [141]3 years ago

5 0

The first one the third one and the fifth one hope this helped

abruzzese [7]3 years ago

3 0

Answer:

The answers are options 1, 2, 3, and 5.

Explanation:

Hello!

Let's solve this!

When an experiment is carried out and less product is obtained, it may be because:

There are impurities in the substances.

The reaction might not go to completion.

The amount of one of the reactants might be measured incorrectly.

The expected amount might be based on measurements with error.

We conclude that the answers are options 1, 2, 3, and 5.

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3 years ago

A 3.50 g sample of an unknown compound containing only C , H , and O combusts in an oxygen‑rich environment. When the products h

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Explanation:

First, calculate the moles of $CO_{2}$ using ideal gas equation as follows.

PV = nRT

or, n = $\frac{PV}{RT}$

= $\frac{1 atm \times 4.41 ml}{0.0821 Latm/mol K \times 293 K}$ (as 1 bar = 1 atm (approx))

= 0.183 mol

As, Density = $\frac{mass}{volume}$

Hence, mass of water will be as follows.

Density = $\frac{mass}{volume}$

0.998 g/ml = $\frac{mass}{3.26 ml}$

mass = 3.25 g

Similarly, calculate the moles of water as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{3.25 g}{18.02 g/mol}$

= 0.180 mol

Moles of hydrogen = $0.180 \times 2$ = 0.36 mol

Now, mass of carbon will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

0.183 mol = $\frac{mass}{12 g/mol}$

= 2.19 g

Therefore, mass of oxygen will be as follows.

Mass of O = mass of sample - (mass of C + mass of H)

= 3.50 g - (2.19 g + 0.36 g)

= 0.95 g

Therefore, moles of oxygen will be as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{0.95 g}{16 g/mol}$

= 0.059 mol

Now, diving number of moles of each element of the compound by smallest no. of moles as follows.

C H O

No. of moles: 0.183 0.36 0.059

On dividing: 3.1 6.1 1

Therefore, empirical formula of the given compound is $C_{3}H_{6}O$ .

Thus, we can conclude that empirical formula of the given compound is $C_{3}H_{6}O$ .

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