<u>Answer:</u> The molality of potassium hydroxide solution is 0.608 m
<u>Explanation:</u>
We are given:
3.301 mass % of potassium hydroxide solution.
This means that 3.301 grams of potassium hydroxide is present in 100 grams of solution
Mass of solvent = Mass of solution - Mass of solute (KOH)
Mass of solvent = (100 - 3.301) g = 96.699 g
To calculate the molality of solution, we use the equation:
Where,
= Given mass of solute (KOH) = 3.301 g
= Molar mass of solute (KOH) = 56.1 g/mol
= Mass of solvent = 96.699 g
Putting values in above equation, we get:
Hence, the molality of potassium hydroxide solution is 0.608 m
They are different by a phase shift of pi/2
1. No two elements have the same kind of atom.
Explanation:
The correct of all statement is that no two elements have the same kind of atoms.
Every element have different atoms.
- Over a hundred elements have been identified by scientists.
- Atoms are the smallest particles that takes part in chemical reactions.
- Elements are distinct substances that cannot be split.
- Atoms of all elements do not have the same mass and volume. Isotopes are examples.
- It is true that atoms of all elements have different masses and also different volumes.
- The volume of an atom is function of the number of electrons it contains.
learn more:
Dalton model of the atom brainly.com/question/1979129
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According to this formula :
㏑[A] /[Ao] = - Kt
when we have Ao = 0.3 m
and K =0.46 s^-1
t = 20min = 0.2 x 60 =12 s
So by substitution :
㏑[A] / 0.3 = - 0.46 * 12
㏑[A] / 0.3 = - 5.52
by taking e^x for both side of the equation we can get [A]
∴[A] = 0.0012 mol dm^-3
Answer:
Explanation:
Cu²⁺ + 2e⁻ → Cu ( copper gets reduced )
Cu → Cu²⁺ + 2e⁻ ( copper gets oxidized )
Oxidation:
Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
Consider the following reactions.
4KI + 2CuCl₂ → 2CuI + I₂ + 4KCl
the oxidation state of copper is changed from +2 to +1 so copper get reduced.
CO + H₂O → CO₂ + H₂
the oxidation state of carbon is +2 on reactant side and on product side it becomes +4 so carbon get oxidized.
Na₂CO₃ + H₃PO₄ → Na₂HPO₄ + CO₂ + H₂O
The oxidation state of carbon on reactant side is +4. while on product side is also +4 so it neither oxidized nor reduced.
H₂S + 2NaOH → Na₂S + 2H₂O
The oxidation sate of sulfur is -2 on reactant side and in product side it is also -2 so it neither oxidized nor reduced.
