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3 years ago

What will be produced at the cathode during the electrolysis of an aqueous solution of magnesium iodide, MgI2

Chemistry

1 answer:

natita [175]3 years ago

6 0

Answer:

At the cathode during the electrolysis of an aqueous solution of magnesium iodide, MgI2 , 2I−(aq) is produced

Explanation:

At cathode, reduction reaction takes place.

The dissociation of MgI2 in aqueous solution is Mg2+(aq) and 2I−(aq)

Here, the Iodine reduces to 2I−(aq) from state of 0 (MgI2) to state of -1 (2I−(aq))

Hence, at the cathode during the electrolysis of an aqueous solution of magnesium iodide, MgI2 , 2I−(aq) is produced

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Once again rocks and/or minerals

sergeinik [125]

Answer:

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Explanation:

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3 years ago

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Find the mass of an object on earth if its weight is 100N

Kaylis [27]

Answer:

0.19 kg

8 N

1.964 N

58.86 N

15 kg

Explanation:

w = Weight of an object = 100 N

g = Acceleration due to gravity of Earth = $9.81\ \text{m/s}^2$

m = Mass of an object

Weight is given by

$w=mg\\\Rightarrow m=\dfrac{w}{g}\\\Rightarrow m=\dfrac{100}{9.81}\\\Rightarrow m=10.19\ \text{kg}$

Mass of the object is 0.19 kg

g = $2\ \text{m/s}^2$

m = 4 kg

$w=mg\\\Rightarrow w=4\times 2\\\Rightarrow w=8\ \text{N}$

Weight of the object is 8 N.

m = 200 g

$w=mg\\\Rightarrow w=0.2\times 9.81\\\Rightarrow w=1.964\ \text{N}$

Weight of the object is 1.964 N.

Weight is the force the Earth exerts on an object which is on the surface of the Earth.

m = 6kg

$w=mg\\\Rightarrow w=6\times 9.81\\\Rightarrow w=58.86\ \text{N}$

The force of the object is 58.86 N.

w = 150 N

a = g = $10\ \text{m/s}^2$

$m=\dfrac{w}{g}\\\Rightarrow m=\dfrac{150}{10}\\\Rightarrow m=15\ \text{kg}$

Mass of the object is 15 kg.

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3 years ago

In the following reaction, when a surplus of hydrazine (N2H4) reacts with 20 liters of hydrogen peroxide (H2O2), how many liters

Ghella [55]

The answer is 10 liters.

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2 years ago

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H is a chemical _____ and H2 is a chemical ______

jeyben [28]

I don’t know if I am right but I think it is symbol for the first one and the second one is formula.

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2 years ago

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Part 1. A chemist reacted 18.0 liters of F2 gas with NaCl in the laboratory to form Cl2 gas and NaF. Use the ideal gas law equat

Alika [10]

Answer:

Part 1

The mass of the NaCl that reacted with F₂ at 290.K and 1.5 atm is approximately 132.6 gams

Part 2

The mass of NaCl that can react with the same volume of gas at STP is approximately 93.77 grams

Explanation:

Part 1

The volume of F₂ gas in the reaction, V = 18.0 liters

The ideal gas equation is P·V = n·R·T

∴ n = P·V/(R·T)

The pressure, P = 1.5 atm

The temperature, T = 290 K

The universal gas constant, R = 0.0820573 L·atm/(mol·K)

∴ n = 1.5×18/(0.0820573 × 290) ≈ 1.134615

The number of moles of F₂ in the reaction n ≈ 1.134615 moles

The chemical reaction is given as follows;

F₂ + 2NaCl → Cl₂ + 2NaF

1 mole of F₂ reacts with 2 moles of NaCl

Therefore;

1.134615 moles of F₂ reacted with 2 × 1.134615 moles ≈ 2.26923 moles of NaCl

1 mole of NaCl = The molar mass of NaCl, MM = 58.44 g/mol

The mass, of 2.26923 moles of NaCl, m = Number of moles × MM

∴ m ≈ 2.26923 moles × 58.44 g/mol ≈ 132.6 grams

The mass of the NaCl ≈ 132.6 gams

Part 2

The volume occupied by 1 mole of all gases at STP = 22.4 l/mole

Therefore, the number of moles of F₂ in 18.0 L of F₂ = 18.0 L/(22.4 L/mole) ≈ 0.804 moles

Therefore;

The number of moles of NaCl, in the reaction n = 2 × The number of moles of F₂ ≈ 2×0.804 moles = 1.608 moles

The number of moles of NaCl, in the reaction n ≈ 1.608 moles

The mass of NaCl in the reaction, m = n × MM

∴ m ≈ 1.608 moles × 58.44 g/mol ≈ 93.97 grams

The mass of NaCl that can react with the same volume of gas at STP ≈ 93.77 grams

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3 years ago