Answer:
The chemical equilibrium of the system will be unaffected. The chemical equilibrium of the system will shift to the right to favor the forward reaction. The chemical equilibrium of the system will shift to the left to favor the reverse reaction. (I hope this helped!!)
Kepler did not study the speed of the planets, rather, he studied how the planets move in the solar system. He proposed three laws. As a summary, he described that the planets move around the sun in the shape of an ellipse (orbit), and the Sun being one of the foci. Then, he proposed the period for the planet to complete one revolution around the Sun.
On the other hand, Newton studied the forces acting on the planet (or any object in space) that explain how the planets move around the solar system as described by Kepler. Also, Kepler's observations only apply to planets and not the moons or satellites. Thus, Kepler only made laws from observations, while Newton based it from underlying principles that led him to mathematical equations such as the law of universal gravitation.
Equation for Half life :
A = a(0.5)^(t/h)
A is current amount, "a" is initial amount, h is halflife, t is time
5 = 40(0.5)^(t/1.3x10^9)
5/40 = (0.5)^(t/1.3x10^9)
take the log of both sides , power rule
Log(5/40) = (t/1.3x10^9) * Log(0.5)
(1.3x10^9) * Log(5/40) / Log(0.5) = t
3.9x10^9 years = t
And if you think about what a half life is, the time it take for the amount to reduce to half.
40/2 = 20
20/2 = 10
10/2 = 5
It went through 3 half-lifes
3 * 1.3x10^9 = 3.9x10^9 years
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.
