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3 years ago

What would be the oxidation and reduction half reactions for this equation?

Chemistry

1 answer:

Sedbober [7]3 years ago

5 0

Answer:

Fe(s) → Fe²⁺(aq) + 2e⁻ OXIDATION

Mg²⁺(aq) + 2e⁻ → Mg(s) REDUCTION

Explanation:

The redox reaction is: MgCl₂(aq) + Fe(s) → FeCl₂(aq) + Mg(s)

We need to know that elements in ground state have 0 as the oxidation state.

Iron in the reactants, and Mg in the products

In the magnessium chloride, the Mg acts with+2, so the oxidation state has decreased → REDUCTION

In the iron(II) chloride, the Fe acts with +2, so the oxidation statehas increased → OXIDATION

The half reactions are:

Fe(s) → Fe²⁺(aq) + 2e⁻ OXIDATION

Mg²⁺(aq) + 2e⁻ → Mg(s) REDUCTION

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Helium has a density of 1.79 x 10-4 g/mL at standard temperature and pressure. A balloon has a volume of 6.3 liters. Calculate t

LenaWriter [7]

The density of He is 1.79 x 10⁻⁴ g/mL
In other words in 1 mL there's 1.79 x 10⁻⁴ g of He.
To fill a volume of 6.3 L the mass of He required
= 1.79 x 10⁻⁴ g/mL * 6300 mL
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3 years ago

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The partial pressure of CO2 gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO2 gas (in g) will be released f

frutty [35]

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. We can calculate the concentration of CO₂ using Henry's law.

$C = k \times P = \frac{1.65 \times 10^{-3} M }{atm} \times 4.60 atm = 7.59 \times 10^{-3} M$

We can calculate the mass of CO₂ in 1.1 L considering its molar mass is 44.01 g/mol.

$\frac{7.59 \times 10^{-3} mol}{L} \times 1.1 L \times \frac{44.01 g}{mol} = 0.367 g$

Now, we will repeat the same procedure for a partial pressure of 1.28 atm.

$C = k \times P = \frac{1.65 \times 10^{-3} M }{atm} \times 1.28 atm = 2.11 \times 10^{-3} M$

$\frac{2.11 \times 10^{-3} mol}{L} \times 1.1 L \times \frac{44.01 g}{mol} = 0.102 g$

The mass of CO₂ released will be equal to the difference in the masses at the different pressures.

$m = 0.367 g - 0.102 g = 0.265 g$

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

Learn more: brainly.com/question/18987224

<em>The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO₂ gas (in g) will be released from 1.1 L of the carbonated water when the partial pressure of CO2 is lowered to 1.28 atm? At 25 ºC, the Henry’s law constant for CO₂ dissolved in water is 1.65 x 10⁻³ M/atm, and the density of water is 1.0 g/cm³.</em>

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2 years ago

A rocket will rotate about a single point called the _?

Leokris [45]

Answer:

Center of gravity.

Explanation:

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3 years ago

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Which of the analogies below is a location-based analogy?

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D is the correct answrt

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3 years ago

A 118-ml flask is evacuated and found to have a mass of 97.129 g. when the flask is filled with 768 torr of helium gas at 35 ?c,

Inessa05 [86]

The full question asks to decide whether the gas was a specific gas. That part is missing in your question. You need to decide whether the gas in the flask is pure helium.

To decide it you can find the molar mass of the gas in the flask, using the ideal gas equation pV = nRT, and then compare with the molar mass of the He.

From pV = nRT you can find n, after that using the mass of gass in the flask you use MM = mass/moles.

1) From pV = nRT, n = pV / RT

Data:
V = 118 ml = 0.118 liter
R = 0.082 atm*liter/mol*K
p = 768 torr * 1 atm / 760 torr = 1.0105 atm
T = 35 + 273.15 = 308.15 K

n = 1.015 atm * 0.118 liter / [ 0.082 atm*liter/K*mol * 308.15K] =0.00472 mol

mass of gas = mass of the fask with the gas - mass of the flasl evacuated = 97.171 g - 97.129 g = 0.042

=> MM = mass/n = 0.042 / 0.00472 = 8.90 g/mol

Now from a periodic table or a table you get that the molar mass of He is 4g/mol

So the numbers say that this gas is not pure helium , because its molar mass is more than double of the molar mass of helium gas.

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3 years ago