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2 years ago

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What is the oxidation number of manganese (Mn) in potassium permanganate (KMnO4)?

2 answers:

irina [24]2 years ago

8 0

Answer:

+7

Explanation:

Within every molecule, there is always an oxidation number assigned to every individual atom within the molecule. Hydrogen will always have an oxidation number of +1 (because it only has one electron to give, and when it gives that electron, it has a charge of +1).

Elements in group 1 always have an oxidation number of +1, and elements in group 16 always have an oxidation number of -2. Using this information we can calculate the oxidation number of Mn in the molecule KMnO₄.

Overall the molecule is neutral, so all the oxidation numbers would add up to zero:

K + Mn + O₄ = 0

(+1) + Mn + 4(-2) = 0

(+1) + Mn + (-8) = 0

Mn + (-7) = 0

Mn = +7

andrezito [222]2 years ago

7 0

What is the oxidation number of manganese (Mn) in potassium permanganate (KMnO4)?

+3
+5
+6
+7

The oxidation number of manganese (Mn) in potassium permanganate (KMnO4) is +5.

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3 years ago

Dihydrogen dioxide decomposes into water and oxygen gas. Calculate the amounts requested if 1.34 moles of dihydrogen dioxide rea

a_sh-v [17]

Answers:

<em><u>a. Moles of oxygen formed:</u></em><u> 0.670 mol</u>

<em><u>b. Moles of water formed:</u></em><u> 1.34 mol</u>

<em><u>c. Mass of water formed:</u></em><u> 24.1 g</u>

<em><u>d. Mass of oxygen formed:</u></em><u> 21.4 g</u>

<u></u>

Explanation:

Dihdyrogen dioxide is the chemical name for a compound made of two hydrogen atoms and two oxide atoms, i.e. H₂O₂, which is also known as hydrogen peroxide or oxygenated water.

The decomposition reaction of dihydrogen dioxide into water and oxygen gas is represented by the balanced chemical equation:

$2H_2O_2(l)\rightarrow 2H_2O(l)+O_2(g)$

The mole ratios derived from that balanced chemical equation are:

2 mol H₂O₂ : 2 mol H₂O : 1 mol O₂

<em><u>a. Moles of oxygen formed</u></em>

Set the proportion using the theoretical mole ratio of H₂O₂ to O₂ and the amount of moles of dyhydrogen dioxide that react:

$2\text{ mol }H_2O_2/1\text{ mol }O_2=1.34\text{ mol }H_2O_2/x$

When you solve for x, you get:

x = 1.34 mol H₂O₂ × 1 mol O₂ / 2 mol H₂O₂ = 0.670 mol O₂

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<em><u>b. Moles of water formed</u></em>

Set the proportion using the theoretical mole ratio of H₂O₂ to H₂O and the amount of moles of dyhydrogen dioxide that react:

$2\text{ mol }H_2O_2/2\text{ mol }H_2O=1.34\text{ mol }H_2O_2/x$

When you solve for x, you get:

x = 1.34 mol H₂O₂ × 2 mol H₂O / 2 mol H₂O₂ = 1.34 mol H₂O

<em><u>c. Mass of water formed</u></em>

Using the number of moles of water calculated in the part b., you calculate the mass of water formed, in grams, using the molar mass of water:

Molar mass of water = 18.015 g/mol

Number of moles = mass in grams / molar mass

⇒ mass in grams = number of moles × molar mass

⇒ mass in grams = 1.34 mol × 18.015 g/mol = 24.1 g

<em><u>d. Mass of oxygen formed</u></em>

Using the number of moles of oxygen determined in the part a., you calculate the mass in grams using the molar mass of O₂.

Molar mass of O₂ = 32.00 g/mol

mass = molar mass × number of moles

mass = 32.00 g/mol × 0.670 mol = 21.4 g.

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What is the binding energy of a mole of nuclei with a mass defect of 0.00084<br> kg/mol?

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