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aivan3 [116]
3 years ago
15

When the nail was copper-plated in Part III of the lab, what element do you think was reduced? Explain your answer in complete s

entences.
Chemistry
1 answer:
Jet001 [13]3 years ago
6 0
Reduction <span>always results in a lowering of the oxidation number. The reaction of the system above is written as:

</span><span>Cu2+(aq) + Fe(s) --> Cu(s) + Fe2+(aq) 
</span>
From the reaction, we see that copper goes from the +2 to a neutral charge. Lowering of the oxidation number happens so this is the element that is being reduced.
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Answer:

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3 years ago
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Lead forms two compounds with oxygen. One contains 2.98g of lead and 0.461g of oxygen. The other contains 9.89g of lead and 0.76
aliina [53]

The given question is incomplete, the complete question is:

Lead forms two compounds with oxygen. One contains 2.98g of lead and 0.461g of oxygen. The other contains 9.89g of lead and 0.763g of oxygen. For a given mass of oxygen, what is the lowest whole-number mass ratio of lead in the two compounds that combines with a given mass of oxygen?

Answer:

The lowest whole-number mass ratio in the two compounds is 1:2.

Explanation:

There is a need to find the mole ratio between lead and oxygen atoms in order to find the whole-number mass ratio of lead in the two compounds. In the first compound, the given mass of lead is 2.98 grams, the molar mass of lead is 207.2 gram per mole.  

The no. of moles can be determined by using the formula,  

moles = mass/molecular mass

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The mass of oxygen in the compound I is 0.461 grams, the molecular mass of oxygen is 16 gram per mol.  

moles = 0.461 g /16 g/mol

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On the other hand, in the compound II, the mass of lead given is 9.89 grams, therefore, the moles of lead in compound II is,  

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The mass of oxygen given in compound II is 0.763 grams, the moles of oxygen present in the compound II is,  

moles = 0.763 g / 16 g

= 0.0477 moles

The ratio between the lead and oxygen in the compound II is, 0.0477 moles lead /0.0477 moles oxygen = 1:1

Hence, of the two compounds, the lowest ratio is found in the compound I, that is, 1:2.  

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