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3 years ago

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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A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the

baherus [9]

Answer: 1. $9.08\times 10^{-6}$ moles

2. 90 mg

Explanation:

$O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus $4.54 \times 10^{-6}$ moles of ozone is removed by = $\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6}$ moles of sodium iodide.

Thus $9.08\times 10^{-6}$ moles of sodium iodide are needed to remove $4.54\times 10^{-6}$ moles of $O_3$

2. $\text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles$

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by = $\frac{2}{1}\times 0.0003=0.0006$ moles of sodium iodide.

Mass of sodium iodide= $moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg$ (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of $O_3$ .

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3 years ago

What best describes how the behavior of rocks changes when they become deeply buried and placed under high levels of heat and pr

Debora [2.8K]

Answer:

They become ductile and deform plastically

Explanation:

When rocks are buried by the materials up to a greater depth, then the confining pressure increases significantly. This results in the ductile behavior of the rocks at such depth. These rocks are present in the ductile region where the depth is about more than 20 to 30 km. Here the rocks are subjected to extremely high pressure and temperature conditions, which favors the transformation of rocks into more higher-grade metamorphic rocks. It is also enhanced due to the geothermal gradient.

Under such high pressure and temperature, the rocks show the behavior of plasticity, where the rocks undergo bending, buckling as well as they tend to flow, and there occurs low strain rate, resulting in the permanent deformation of rocks.

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3 years ago

A composition of reflections over parallel lines is the same as a __________.

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Answer : Option A) Translation

Explanation : A composition of reflections over parallel lines is the same as a Translation.

To identify if the composition of reflections over parallel lines are same as translation or not?

We can check using a picture of some shape in the plane. Place the picture on the right side of two vertical parallel. Now, we can see the reflected the shape over the nearest parallel line, then check the reflection over the other parallel line. We see that the shape winds up in the same orientation, like it was just shifted over to the right. Hence, it is translation.

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3 years ago

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3 years ago

You need 455 ml of a 75% alcohol solution. on hand, you have a 25% alcohol mixture. you also have a 90% alcohol mixture. how muc

marishachu [46]

X ml - 25% alcohol mixture
y ml - 90% alcohol mixture

x+y = 455

0.25x ml alcohol in x ml of 25% alcohol mixture
0.9y ml alcohol in y ml of 90% alcohol mixture
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0.25x+0.9y=341.25

System of equations:

x+y = 455 /*(-0.25) ------> -0.25x-0.25y = -0.25*455
0.25x+0.9y=341.25

-0.25x-0.25y=-113.75
0.25x+0.9y=341.25 Add both equations

0.25x+0.9y-0.25x-0.25y=341.25-113.75
0.65y =227.5
y=227.5/0.65 = 350 ml of 90% alcohol mixture

x+y=455
x+355=455
x=100 ml of 25% alcohol mixture

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3 years ago