In order to identify the reducing agent let us check the oxidation states
I2 + Ti+4 ----> Ti + IO3- reducing agent
I2 : oxidation state of I = 0
Ti^+4 = oxidation state of Ti = +4
Ti : oxidation state = 0
IO3^-1 , oxidation state of Iodine = +5
So here the Ti undergoes reduction hence it is oxidizing agent
Where iodine undergoes oxidation hence it is reducing agent
Swarm behavior x))))))))))))))))))
Answer:
- <em>The vapor pressure of the solution at 25°C is </em><u><em>23.6 mmHg</em></u>
Explanation:
<u>1) Data:</u>
- Mass of solute, sucroese, m₁ = 40.0 g
- Mass of solvent, water, m₂ = 250. g
- Vapor pressure of solution, p = ?
- Vapor pressure of pure water, p⁰ = 23.76 mm Hg
<u>2) Principles and formulae:</u>
- Raoult's law states that the vapor pressure of a solution is equal to the mole fraction of the solvent times the vapor pressure of the pure liquid.
p = X p⁰
<u>3) Solution:</u>
a) <u>Mole fraction of the solvent (water)</u>
- X = number of moles of solvent / number of moles of solution
- number of moles of solvent, n₁ = mass in grams / molar mass
n₁ = 250. g / 18.015 g/mol = 13.88 moles
- number of moles of solute, n₂ = mass in grams / molar mass
n₂ = 40.0 g / 342,3 g/mol = 0.12 moles
- total number of moles, n₁ + n₂ = 13.88 moles + 0.12 moles = 14.0 moles
- moles fraction of water, X = 13.88 moles / 14.0 moles = 0.99
b) <u>Vapor pressure of the solution</u>:
- p = p⁰ X = 23.76 mmHg × 0.99 = 23.56 mm Hg
Rounding to three significant figures: 23.6 mm Hg.
Answer:
200 grams of sodium hydroxide produce 595 grams of sodium sulfate if we have an excess of sulphuric acid
Explanation:
2 NaOH + H2SO4 → 2 H20 + Na S04
From the above chemical equation we can state that
2 moles of sodium hydroxide reacts with 1 mole of sulfuric acid to produce 1 mole of sodium sulfate.
Now if we start with 200 grams of sodium hydroxide and you have an excess of sulfuric acid then the amount in grams of sodium sulfate will be formed
2 moles of sodium hydroxide produce 1 mole of sodium sulfate.
In term of mass, we can say that
40 grams of sodium hydroxide produce 119grams of sodium sulfate
1 grams of sodium hydroxide produce (119/40) grams of sodium sulfate
then 200 grams of sodium hydroxide produce ((119*200)/40) grams of sodium sulfate
or 200 grams of sodium hydroxide produce 595 grams of sodium sulfate
The answer is A ( pure water has no free ions)