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shepuryov [24]
3 years ago
13

Very unsure on my answers for these 2 questions. I really have no idea what I'm doing. Could someone please explain? Will give m

edal and maybe become a fan if someone would just help me understand! Please?
1. Which of the following energy yields is likely to have come from a fission or fusion reaction?
1.0 × 10^2 kJ/mol

2.5 × 10^2 kJ/mol

1.2 × 10^3 kJ/mol

*This one??? 1.4 × 10^11 kJ/mol


2. In a nuclear reaction, a change in mass of 1.0 × 104 g occurs. What is the corresponding change in energy? Use the formula E = mc2.
9.0 × 10^3 kJ

*My best guess... 9.0 × 10^6 kJ

9.0 × 10^9 kJ

9.0 × 10^12 k
Chemistry
1 answer:
murzikaleks [220]3 years ago
6 0
The answers are the following:
1. <span>1.4 x 10^11 kJ/mol
2. </span><span>9.0 × 10^6 kJ

Therefor I conclude that your answers are correct. </span>
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Answer:

New Moon, First Quarter, Full Moon, Last Quarter, Waxing Crescent, Waxing Gibbous, Waning Gibbous, and Waning Crescent.

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As a result of high winds and water from a storm surge, homes, businesses, and crops may be destroyed or damaged, public infrastructure may also be compromised, and people may suffer injuries or loss of life.
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2 years ago
The dissociation of sulfurous acid (H2SO3) in aqueous solution occurs as follows:
aksik [14]

Answer:

The [SO₃²⁻]

Explanation:

From the first dissociation of sulfurous acid we have:

                         H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)

At equilibrium:  0.50M - x          x            x

The equilibrium constant (Ka₁) is:

K_{a1} = \frac{[H^{+}] [HSO_{3}^{-}]}{[H_{2}SO_{3}]} = \frac{x\cdot x}{0.5 - x} = \frac {x^{2}}{0.5 -x}

With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:

[HSO_{3}^{-}] = [H^{+}] = 7.94 \cdot 10^{-2}M

Similarly, from the second dissociation of sulfurous acid we have:

                              HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)

At equilibrium:  7.94x10⁻²M - x          x            x

The equilibrium constant (Ka₂) is:  

K_{a2} = \frac{[H^{+}] [SO_{3}^{2-}]}{[HSO_{3}^{-}]} = \frac{x^{2}}{7.94 \cdot 10^{-2} - x}  

Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:

[SO_{3}^{2-}] = [H^{+}] = 7.07 \cdot 10^{-5}M

Therefore, the final concentrations are:

[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M

[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M

[SO₃²⁻] = 7.07x10⁻⁵M

[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M

So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.

I hope it helps you!

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Only then will the <em>charges cancel</em> and give a neutral atom.


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