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raketka [301]
3 years ago
14

(part 1 of 3) Copper reacts with silver nitrate through a single replacement. If 1.29 g of silver are produced from the reaction

, how much copper(II) nitrate is also produced? Answer in units of mol. (part 2 of 3) How much Cu is required in this reaction? Answer in units of mol. (part 3 of 3) 1.0 points How much AgNO3 is required in this reaction? Answer in units of mol.
Chemistry
1 answer:
ale4655 [162]3 years ago
5 0

Answer:

See explanation.

Explanation:

Hello there!

In this case, according to the described chemical reaction, we first write the corresponding equation to obtain:

Cu+2AgNO_3\rightarrow 2Ag+Cu(NO_3)_2

Thus, we proceed as follows:

Part 1 of 3: here, since the molar mass of silver and copper (II) nitrate are 107.87 and 187.55 g/mol respectively, and the mole ratio of the former to the latter is 2:1, we can set up the following stoichiometric expression:

m_{Cu(NO_3)_2}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu(NO_3)_2}{2molAg}*\frac{187.55gCu(NO_3)_2}{1molCu(NO_3)_2}   \\\\m_{Cu(NO_3)_2}=1.12gCu(NO_3)_2

Part 2 of 3: here, the molar mass of copper is 63.55 g/mol and the mole ratio of silver to copper is 2:1, the mass of the former that was used to start the reaction was:

m_{Cu}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu}{2molAg}*\frac{63.55gCu)_2}{1molCu}   \\\\m_{Cu}=0.380gCu

Part 3 of 3: here, the molar mass of silver nitrate is 169.87 g/mol and their mole ratio 2:2, thus, the mass of initial silver nitrate is:

m_{AgNO_3}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{2molAgNO_3}{2molAg}*\frac{169.87gAgNO_3}{1molAgNO_3}   \\\\m_{AgNO_3}=2.03gAgNO_3

Best regards!

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Considering the definition of atomic mass, isotopes and atomic mass of an element, sb121 has a percent natural abundance of 0.5726 or 57.26% and sb123 has a percent natural abundance of 0.4284 or 42.84%.

<h3>Definition of atomic mass</h3>

The atomic mass is obtained by adding the number of protons and neutrons in a given nucleus of a chemical element.

<h3>Definition of isotope</h3>

Isotopes are the chemical elements in which atomic numbers are the same, but the number of neutrons is different.

<h3>Definition of atomic mass</h3>

The atomic mass of an element is the weighted average mass of its natural isotopes.

This is, the atomic masses of elements are usually calculated as the weighted average of the masses of the different isotopes of each element, considering the relative abundance of each of them.

<h3>Percent natural abundance of each isotope</h3>

In this case, antimony has two naturally occuring isotopes, sb121 and sb123. You know:

  • sb121 has an atomic mass of 120.9038 u.
  • sb121 has a percent natural abundance of x.
  • sb123 has an atomic mass of 122.9042 u.
  • sb123 has a percent natural abundance of 1 -x (or, what is the same, the abundance is 100% - x%, since both isotopes form 100% of the element.)
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The average mass of antimony is expressed as:

121.7601 u= 120.9038 u x + 122.9042 u× (1 -x)

Solving:

121.7601 u= 120.9038 u x + 122.9042 u - 122.9042 u x

121.7601 u - 122.9042 u= 120.9038 u x - 122.9042 u x

(-1.1441 u)= (-2.0014) x

(-1.1441 u)÷ (-2.0014)= x

<u><em>0.5726= x or 57.26%</em></u>

So, 1 -x= 1- 0.5716 → <u><em>1-x= 0.4284 or 42.84%</em></u>

<u><em /></u>

Finally, sb121 has a percent natural abundance of 0.5726 or 57.26% and sb123 has a percent natural abundance of 0.4284 or 42.84%.

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