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Natalija [7]
3 years ago
10

A lead atom has a mass of 3.14 x 10 to the negative 22nd g.How

Chemistry
1 answer:
frozen [14]3 years ago
4 0

Explanation:

The given data is as follows.

       Mass of a lead atom = 3.14 \times 10^{-22}

       Volume = 2.00 cm^{3}

        Density = 11.3 g/cm^{3}

As it is mentioned that 1 cubic centimeter contains 11.3 grams of lead.

So, in 2 cubic centimeter there will be 2 \times 11.3 g = 22.6 g of lead atoms.

One lead atom has a mass of 3.14 \times 10^{-22}. Therefore, number of atoms present in 22.6 g of lead will be as follows.

                \frac{22.6 g}{3.14 \times 10^{-22}}

                  = 7.197 \times 10^{22} atoms

Thus, we can conclude that there are 7.197 \times 10^{22} atoms of lead are present.

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Part 1: What is the final volume in milliliters when 0.730 L of a 44.8 % (m/v) solution is diluted to 23.3 % (m/v)?
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part 1 : the final volume : 1.404 L

part 2 : the initial concentration : 4.06 M

<h3>Further explanation </h3>

Dilution is the process of adding a solvent to get a more dilute solution.

The moles(n) before and after dilution are the same.

Can be formulated :

M₁V₁=M₂V₂

M₁ = Molarity of the solution before dilution  

V₁ = volume of the solution before dilution  

M₂ = Molarity of the solution after dilution  

V₂ = Molarity volume of the solution after dilution

part 1 :

M₁=44.8%

V₁=0.73 L

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\tt V_2=\dfrac{M_1.V_1}{M_2}\\\\V_2=\dfrac{44.8\times 0.73}{23.3}\\\\V_2=1.404~L

part 2 :

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\tt M_1=\dfrac{M_2.V_2}{V_1}\\\\M_1=\dfrac{2\times 1.5}{0.739}\\\\M_1=4.06

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3 years ago
The measurement of grams of solute per one million grams of solution is known as
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5 0
4 years ago
Electrochemistry - Equilibrium
Ipatiy [6.2K]

Answer:

Explanation:

The relation between equilibrium constant and Ecell is given below .

E⁰cell = (RT / nF ) lnK  , F is faraday constant T is 273 + 25 = 298 K

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n = 2

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.25 = (8.314 x 298  lnK) / (2 x 96485 )

lnK = 19.47

K = 2.85 x 10⁸

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Change in free energy Δ G

Δ G ⁰ = nE⁰ F

n = 4

E⁰ = .4 + .83 = 1.23 V

Δ G ⁰= 4 x 1.23 x 96485

= 474706 J / mol

3 )

E⁰cell = (RT / nF ) lnK

n = 2

1.78 = 8.314 x 298  lnK / 2 x 96485

lnK = 138.638

K = 1.62 x 10⁶⁰

8 0
3 years ago
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