The number of mole of ammonium ion, NH₄⁺ in the solution is 0.175 mole
We'll begin by calculating the number of mole of NH₄NO₃ in the solution. This can be obtained as follow:
Volume = 125 mL = 125 / 1000 = 0.125 L
Molarity = 1.40 M
<h3>Mole of NH₄NO₃ =? </h3>
Mole = Molarity x Volume
Mole of NH₄NO₃ = 1.40 × 0.125
<h3>Mole of NH₄NO₃ = 0.175 mole</h3>
Finally, we shall determine the number of mole of ammonium ion, NH₄⁺ in the solution. This can be obtained as follow:
NH₄NO₃(aq) —> NH₄⁺(aq) + NO₃¯(aq)
From the balanced equation above,
1 mole of NH₄NO₃ contains 1 mole of NH₄⁺
Therefore,
0.175 mole of NH₄NO₃ will also contain 0.175 mole of NH₄⁺
Thus, the number of mole of ammonium ion, NH₄⁺ in the solution is 0.175 mole
Learn more: brainly.com/question/25469095
<span>Answer: 0.00649M
The question is incomplete,
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<span>You are told that the first ionization of the sulfuric acid is complete and the second ionization of the sulfuric acid has a constant Ka₂ = 0.012
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With that you can solve the question following these steps"
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<span>1) First ionization:
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<span>
H₂SO₄(aq) --> H⁺ (aq) + HSO₄⁻ (aq)
Under the fully ionization assumption the concentration of HSO4- is the same of the acid = 0.01 M
2) Second ionization
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<span>HSO₄⁻ (aq) ⇄ H⁺ + SO₄²⁻ with a Ka₂ = 0.012
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<span>Do the mass balance:
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<span><span> HSO₄⁻ (aq) H⁺ SO₄²⁻</span>
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<span /><span /><span> 0.01 M - x x x
</span><span>Ka₂ = [H⁺] [SO₄²⁻] / [HSO₄⁻]</span>
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=> Ka₂ = (x²) / (0.01 - x) = 0.012
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<span>3) Solve the equation:
</span><span>x² = 0.012(0.01 - x) = 0.00012 - 0.012x</span>
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x² + 0.012x - 0.0012 = 0
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<span>Using the quadratic formula: x = 0.00649
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<span>So, the requested concentratioN is [SO₄²⁻] = 0.00649M</span>
Answer:
The correct option is;
d 4400
Explanation:
The given parameters are;
The mass of the ice = 55 g
The Heat of Fusion = 80 cal/g
The Heat of Vaporization = 540 cal/g
The specific heat capacity of water = 1 cal/g
The heat required to melt a given mass of ice = The Heat of Fusion × The mass of the ice
The heat required to melt the 55 g mass of ice = 540 cal/g × 55 g = 29700 cal
The heat required to raise the temperature of a given mass ice (water) = The mass of the ice (water) × The specific heat capacity of the ice (water) × The temperature change
The heat required to raise the temperature of the ice from 0°C to 100°C = 55 × 1 × (100 - 0) = 5,500 cal
The heat required to vaporize a given mass of ice = The Heat of Vaporization × The mass of the ice
The heat required to vaporize the 55 g mass of ice at 100°C = 80 cal/g × 55 g = 4,400 cal
The total heat required to boil 55 g of ice = 29700 cal + 5,500 cal + 4,400 cal = 39,600 cal
However, we note that the heat required to vaporize the 55 g mass of ice at 100°C = 80 cal/g × 55 g = 4,400 cal.
The heat required to vaporize the 55 g mass of ice at 100°C = 4,400 cal
Answer:
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Explanation:
