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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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Determine the oxidation number of sodium in Na202

Olegator [25]

Answer:

Explanation:

Na₂O₂

NOTE: the oxidation number of oxygen is always –2 except in peroxides where it is –1.

Thus, we can obtain the oxidation number of sodium (Na) in Na₂O₂ as illustrated below:

Na₂O₂ = 0 (oxidation number of ground state compound is zero)

2Na + 2O = 0

O = –1

2Na + 2(–1) = 0

2Na – 2 = 0

Collect like terms

2Na = 0 + 2

2Na = 2

Divide both side by 2

Na = 2/2

Na = +1

Thus, the oxidation number of sodium (Na) in Na₂O₂ is +1

5 0

2 years ago

In an electroplating process, copper (ionic charge +2e, atomic weight 63.6 g/mol) is deposited using a current of 10.0 A. What m

salantis [7]

Answer : The mass of copper deposit is, 1.98 grams

Explanation :

First we have to calculate the charge.

Formula used : $Q=I\times t$

where,

Q = charge = ?

I = current = 10 A

t = time = 10 min = 600 sec (1 min = 60 sec)

Now put all the given values in this formula, we get

$Q=10A\times 600s=6000C$

Now we have to calculate the number of atoms deposited.

As, 1 atom require charge to deposited = $2\times (1.6\times 10^{-19})$

Number of atoms deposited = $\frac{(6000)}{2\times(1.6\times 10^{-19})}=1.875\times 10^{22}$ atoms

Now we have to calculate the number of moles deposited.

Number of moles deposited = $\frac{(1.875\times 10^{22})}{(6.022\times 10^{23})}=0.03113$ moles

Now we have to calculate the mass of copper deposited.

1 mole of Copper has mass = 63.5 g

Mass of Copper Deposited = $63.5\times 0.03113 =1.98g$

Therefore, the mass of copper deposit is, 1.98 grams

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