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Stels [109]
3 years ago
5

What is the oxidation state of each element in the species Mn(ClO4)3

Chemistry
2 answers:
Kamila [148]3 years ago
4 0

The oxidation state of the compound Mn (ClO4)3 is to be determined in this problem.

For oxygen, the charge is 2-, the total considering its number of atoms is -24.

Mn has a charge of +3.

TO compute for Mn, we must achieve zero charge overall hence 3 + 3x - 24 = 0 where x is the Cl charge.

Cl charge, x is +7.

zavuch27 [327]3 years ago
3 0
First, we'll calculate the valency of chlorine by using the net charge on the ClO₄ ion.
The ClO₄ ion has a charge of -1, while each oxygen atom exhibits a charge of -2. Therefore,

Cl + 4(-2) = -1
Cl = 7

Next, the valency of Mn is calculated by using the overall charge, which is 0

Mn + 3(-1) = 0
Mn = 3

Therefore, the valencies are:
Mn = 3
Cl = 7
O = -2
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(part 1 of 3) Copper reacts with silver nitrate through a single replacement. If 1.29 g of silver are produced from the reaction
ale4655 [162]

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!

5 0
3 years ago
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Consider the following mechanism for the oxidation of bromide ions by hydrogen peroxide in aqueous acid solution.
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Answer:

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Explanation:

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