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beks73 [17]
3 years ago
14

Calculate the equilibrium constant for the reaction between fe2+(aq) and zn(s) under standard conditions at 25∘c.

Chemistry
1 answer:
Hatshy [7]3 years ago
7 0
Following reaction occurs in the given electrochemical system:
Fe^{2+} + Zn → Fe + Zn^{2+}
Thus, under standard conditions
E(0) = E(0) Fe2+/Fe - E(0) Zn2+/Zn
where, E^{0}Fe2+/Fe = standard reduction potential of Fe2+/Fe = -0.44 v
E^{0}Zn2+/Zn = standard reduction potential of Zn2+/Zn = -0.763 v

E(0) = 0.323 v
now, we know that, ΔG(0) =-nFE(0) ............... (1)
Also, ΔG^{0} = -RTln(K) ............ (2)

On equating and rearranging equation 1 and 2, we get
K = exp( \frac{nFE(0)}{RT} )= exp (\frac{2X96500X0.323}{8.314X298}) = 8.46 x 10^{10}

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What does this weather map symbol represent
Black_prince [1.1K]

Answer:

B

Explanation:

Blue triangles are a cold front and red bowls are warm front

Stationary front is a combination

HOPE I helped :]

5 0
3 years ago
Read 2 more answers
What atom has 5 neutrons 4 protons and 2 electrons
Vilka [71]
<span>the atom that has these characteritics is called beryllium nucleus</span>
6 0
3 years ago
A 20.0-milliliter sample of 0.200 M K2CO3 solution is added to 30.0 milliliters of 0.400 M Ba(NO3)2 solution.
jenyasd209 [6]

Answer:

(B) 0.160 M

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For K_2CO_3 :

Molarity = 0.200 M

Volume = 20.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 20.0×10⁻³ L

Thus, moles of K_2CO_3 :

Moles=0.200 \times {20.0\times 10^{-3}}\ moles

Moles of K_2CO_3 = 0.004 moles

For Ba(NO_3)_2 :

Molarity = 0.400 M

Volume = 30.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume =30.0×10⁻³ L

Thus, moles of Ba(NO_3)_2 :

Moles=0.400\times {30.0\times 10^{-3}}\ moles

Moles of Ba(NO_3)_2  = 0.012 moles

According to the given reaction:

K_2CO_3_{(aq)}+Ba(NO_3)_2_{(aq)}\rightarrow BaCO_3_{(s)}+2KNO_3_{(aq)}

1 mole of potassium carbonate react with 1 mole of barium nitrate

0.004 moles potassium carbonate react with 0.004 mole of barium nitrate

Moles of barium nitrate  = 0.004 moles

Available moles of barium nitrate  =  0.012 moles

Limiting reagent is the one which is present in small amount. Thus, potassium carbonate is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of potassium carbonate gives 1 mole of barium carbonate

Also,

0.004 mole of potassium carbonate gives 0.004 mole of barium carbonate

Mole of barium carbonate = 0.004 moles

Also, consumed barium nitrate = 0.004 moles  (barium ions precipitate with carbonate ions)

Left over moles = 0.012 - 0.004 moles = 0.008 moles

Total volume = 20.0 + 30.0 mL = 50.0 mL = 0.05 L

So, Concentration = 0.008/0.05 M = 0.160 M

<u>(B) is correct.</u>

4 0
3 years ago
Aluminum has a density of 2.70 g/m. Calculate the mass (in grams) of a piece of aluminum having a volume of 382 mL
Natalka [10]

The aluminum has a mass of 1031.4 grams.

8 0
3 years ago
(a) 4.12 x 10^15 atoms U
DochEvi [55]

Answer:

1.63ₓ10⁻⁶ g of U

139.03 g of H

0.385 g of O

141.8 g of Pb

Explanation:

In first place, we need to convert the number of atoms to moles, as we know that 1 mol of anything occupies 6.02×10²³ particles

Therefore:

4.12×10¹⁵ atoms of U . 1 mol / 6.02×10²³ atoms = 6.84×10⁻⁹ moles of U

8.37×10²⁵ atoms of H . 1 mol /6.02×10²³ atoms = 139.03 moles of H

1.45×10²² atoms of O . 1 mol /6.02×10²³ atoms = 0.0241 moles of O

4.12×10²³ atoms of Pb . 1 mol /6.02×10²³ atoms = 0.684 moles of Pb

Moles . Molar mass = Mass (g)

6.84×10⁻⁹ moles of U . 238.03 g/mol = 1.63ₓ10⁻⁶ g of U

139.03 moles of H . 1 g/mol = 139.03 g of H

0.0241 moles of O . 16 g/mol = 0.385 g of O

0.684 moles of Pb . 207.2 g/mol = 141.8 g of Pb

8 0
3 years ago
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