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Afina-wow [57]
4 years ago
13

Lead−206 is the end product of 238u decay. one 206pb atom has a mass of 205.974440 amu. (a) calculate the binding energy per nuc

leon in mev: mev/nucleon
Chemistry
1 answer:
Dovator [93]4 years ago
8 0
The atomic number for Pb is 82
∴ Pb has 82 protons and 206-82 = 14 protons
The actual mass of Pb nuclei is
=(82 × mass of the proton) + (124 × mass of neutron)
=(82× 1.00728) + (124 × 1.008664) amu
= 207.6713 amu
The mass of lead which is given is 205.9744 amu
∴mass defect is
m = 207.6713 - 205.9744 = 1.6969 amu
=1.6969 × 1.66054 × 10⁻²⁷kg
=2.818 × 10⁻²⁷kg
The binding energy is E = mc²
C is the speed of light in vacuum = 2.9979 × 10⁸m/s
∴ E = 2.532 × 10×⁻¹⁰ J/mol
= 2.532 × 10⁻¹⁰ × 6.023 × 10²³ J/mol 
= 1.53811 × 10¹⁴ J/mol

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6. What happens when a piece of sodium is dropped into water?​
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Determine the ratio of amount NaOH reacting per unit amount FeCl3. Enter an integer that represents x mol NaOH/1 mol FeCl3.
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Answer is: the ratio of amount NaOH reacting per unit amount FeCl₃ is 3 mol NaOH /1 mol FeCl₃.

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The water-gas shift reaction plays a central role in the chemical methods for obtaining cleaner fuels from coal: CO(g) + H2O(g)
taurus [48]

<u>Answer:</u> The concentration of carbon dioxide, hydrogen gas, carbon monoxide and water when equilibrium is re-established are 0.362 M, 0.212 M, 0.138 M and 0.138 M respectively.

<u>Explanation:</u>

For the given chemical reaction:

CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)

The expression of K_c for above reaction follows:

K_c=\frac{[CO_2][H_2]}{[CO][H_2O]}         ........(1)

We are given:

[CO]_{eq}=[H_2O]_{eq}=[H_2]_{eq}=0.10M

[CO_2]_{eq}=0.40M

Putting values in above equation, we get:

K_c=\frac{0.40\times 0.10}{010\times 0.10}\\\\K_c=4

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Moles of hydrogen gas = 0.30 mol

Volume of solution = 2.0 L

Putting values in above equation, we get:

\text{Molarity of }H_2=\frac{0.30mol}{2L}=0.15M

When hydrogen gas is added, the concentration of product gets increased. But, by Le-Chatelier's principle, the equilibrium will shift in the direction where concentration of product must decrease, which is in the backward direction.

Concentration of hydrogen gas when equilibrium is re-established = 0.1 + 0.15 = 0.25 M

Now, the equilibrium is shifting to the reactant side. The equation follows:

                      CO(g)+H_2O(g)\rightleftharpoons CO_2(g)+H_2(g)

Initial:              0.1      0.1                 0.4       0.1

At eqllm:       0.1+x   0.1+x           0.4-x      0.25-x

Putting values in expression 1, we get:

4=\frac{(0.25-x)(0.4-x)}{(0.1+x)(0.1+x)}\\\\3x^2+1.45x-0.06=0\\\\x=0.038,-0.522

Neglecting the negative value of 'x'

Calculating the concentrations of the species:

Concentration of carbon dioxide = (0.4 - x) = (0.4 - 0.038) = 0.362 M

Concentration of hydrogen gas = (0.25 - x) = (0.25 - 0.038) = 0.212 M

Concentration of carbon monoxide = (0.1 + x) = (0.1 + 0.038) = 0.138 M

Concentration of water = (0.1 + x) = (0.1 + 0.038) = 0.138 M

Hence, the concentration of carbon dioxide, hydrogen gas, carbon monoxide and water when equilibrium is re-established are 0.362 M, 0.212 M, 0.138 M and 0.138 M respectively.

8 0
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snow_tiger [21]

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4 0
3 years ago
Read 2 more answers
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