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Delvig [45]
3 years ago
12

Write a balanced half-reaction for the reduction of permanganate ion (Mno) to manganese ion Mn?) in acidic aqueous solution. Be

sure to add physical state symbols where appropriate. 0-0 Cb X 5 2
Chemistry
1 answer:
Luden [163]3 years ago
6 0

Explanation:

Permanganate is general name for the chemical compound which containes manganate (VII) ion. Permanganate(VII) ion is strong oxidizing agent as  manganese is in +7 oxidation state and can be easily reduced and oxidize others.

The balanced half reaction for reduction of the permanganate ion, MnO_4^- to manganese ion, Mn^{2+} is shown below:

MnO_4^-_{(aq)}+8H^+_{(aq)}+5e^-\rightarrow Mn^{2+}_{(aq)}+4H_2O_{(l)}

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7 0
4 years ago
Enter your answer in the box provided. How many grams of helium must be added to a balloon containing 6.24 g helium gas to doubl
Archy [21]

Answer : The mass of helium gas added must be 12.48 grams.

Explanation : Given,

Mass of helium (He) gas = 6.24 g

Molar mass of helium = 4 g/mole

First we have to calculate the moles of helium gas.

\text{Moles of }He=\frac{\text{Mass of }He}{\text{Molar mass of }He}=\frac{6.24g}{4g/mole}=1.56moles

Now we have to calculate the moles of helium gas at doubled volume.

According to the Avogadro's law, the volume of gas is directly proportional to the number of moles of gas at same pressure and temperature. That means,

V\propto n

or,

\frac{V_1}{V_2}=\frac{n_1}{n_2}

where,

V_1 = initial volume of gas  = V

V_2 = final volume of gas = 2V

n_1 = initial moles of gas  = 1.56 mole

n_2 = final moles of gas  = ?

Now we put all the given values in this formula, we get

\frac{V}{2V}=\frac{1.56mole}{n_2}

n_2=3.12mole

Now we have to calculate the mass of helium gas at doubled volume.

\text{Mass of }He=\text{Moles of }He\times \text{Molar mass of }He

\text{Mass of }He=3.12mole\times 4g/mole=12.48g

Therefore, the mass of helium gas added must be 12.48 grams.

4 0
3 years ago
What is the volume of a cube with dimensions 11.0 cm × 11.0 cm × 11.0 cm in m3?
tatiyna

Answer:

1.33 × 10⁻³ m³

Explanation:

Step 1: Given data

Side of the cube (d): 11.0 cm

Step 2: Calculate the volume of the cube (V)

We will use the following expression.

V = d³

V = (11.0 cm)³

V = 1.33 × 10³ cm³

Step 3: Convert "V" to m³

We will use the conversion factor 1 m³ = 10⁶ cm³.

1.33 × 10³ cm³ × (1 m³ / 10⁶ cm³) = 1.33 × 10⁻³ m³

7 0
4 years ago
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