Answer:
5 moles of electrons
Explanation:
The balance equation is as follow,
<span> 5 Ag</span>⁺ + Mn⁺²<span> + 4 H</span>₂O →<span> 5 Ag + MnO</span>₄⁻<span> + 8 H</span>⁺
Reduction of Ag:
Ag⁺ + 1 e⁻ → Ag
Or,
5 Ag⁺ + 5 e⁻ → 5 Ag
Oxidation of Mn:
Mn⁺² → MnO₄⁻ + 5 e⁻
Result:
Hence 5 moles of Ag⁺ accepts 5 electrons from 1 mole of Mn⁺².
Answer:
A tritium is produced.
Explanation:
Combining two additional neutrons to the nucleus of the hydrogen atom makes it a tritium, Hydrogen-3.
neutron is designated ¹₀n; this shows a mass number of 1 and no atomic number
Hydrogen-1 is designated as ₁¹H; a mass number of 1 and atomic number of 1. This particle is actually more like a proton.
Combining both:
₁¹H + 2¹₀n → ³₁H
This is a nuclear reaction and in balancing such reaction equation, mass numbers and atomic numbers must be conserved.
Answer:
2 H⁺ + 2e = H₂ ( reduction )
Explanation:
Fe( s ) + 2 CH₃COOH = Fe ( OOCCH₃ ) ₂ + H₂
Fe( s ) = Fe⁺² + 2e ( oxidation )
2 H⁺ + 2e = H₂ ( reduction )
The quantity which is equivalent to the product of the absolute index of refraction of water and the speed of light in water system is the speed of light in vacuum.
<h3>What is the speed of light?</h3>
Speed of light is the rate of speed though the light travels. To find the speed of light in any medium, the following formula is used.
Here, (n) is the index of reaction and (c) is the speed of light in the vacuum. The speed of light in the vacuum is almost equal to the 3.0×10⁸ m/s.
Now the quantity which is equivalent to the product of the absolute index of refraction of water and the speed of light in water has to be find out.
The above formula can be written as,
Here, the product of index of refraction and speed of light is equal to the speed of light in vacuum. This will be true for water as well.
Thus, the quantity which is equivalent to the product of the absolute index of refraction of water and the speed of light in water system is the speed of light in vacuum.
Learn more about the speed of light here;
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Answer:
The freezing point of the solution is - 4.39 °C.
Explanation:
We can solve this problem using the relation:
<em>ΔTf = (Kf)(m),</em>
where, ΔTf is the depression in the freezing point.
Kf is the molal freezing point depression constant of water = -1.86 °C/m,
density of water = 1 g/mL.
<em>So, the mass of 575 mL is 575 g = 0.575 kg.</em>
m is the molality of the solution (m = moles of solute / kg of solvent = (465 g / 342.3 g/mol)/(0.575 kg) = 2.36 m.
<em>∴ ΔTf = (Kf)(m</em>) = (-1.86 °C/m)(2.36 m) = <em>- 4.39 °C.</em>
<em>∵ The freezing point if water is 0.0 °C and it is depressed by - 4.39 °C.</em>
<em>∴ The freezing point of the solution is - 4.39 °C.</em>