Answer:
0.0468 g.
Explanation:
- The decay of radioactive elements obeys first-order kinetics.
- For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).
Where, k is the rate constant of the reaction.
t1/2 is the half-life time of the reaction (t1/2 = 1620 years).
∴ k = ln2/(t1/2) = 0.693/(1620 years) = 4.28 x 10⁻⁴ year⁻¹.
- For first-order reaction: <em>kt = lna/(a-x).</em>
where, k is the rate constant of the reaction (k = 4.28 x 10⁻⁴ year⁻¹).
t is the time of the reaction (t = t1/2 x 8 = 1620 years x 8 = 12960 year).
a is the initial concentration (a = 12.0 g).
(a-x) is the remaining concentration.
∴ kt = lna/(a-x)
(4.28 x 10⁻⁴ year⁻¹)(12960 year) = ln(12)/(a-x).
5.54688 = ln(12)/(a-x).
Taking e for the both sides:
256.34 = (12)/(a-x).
<em>∴ (a-x) = 12/256.34 = 0.0468 g.</em>
by putting to much current through it ?
Answer:
a, b, c, d
Explanation:
Rutherford’ atomic model is based on the gold foil experiment. In this experiment, beam of alpha rays was bombarded on thin gold foil. He observed that:
Most of the alpha particles passed through thin foil without any deflection.
Few alpha particles deflected by an angle of 90o.
Based on observation, Rutherford concluded that majority of the space inside the atom is empty.
He explained defection of few alpha particles by assuming that most of the mass is concentrated at the nucleus and positively charged.
Therefore, among given, the correct statements are:
The atom contains a positively charged nucleus.
Positive charge is condensed in one location within the atom.
The majority of the space inside the atom is empty space
The mass of an atom is concentrated at the nucleus
Therefore, the correct options are:
a, b, c, d
When mixture of NaCl and Al₂(SO₄)₃ is allowed to react with excess NaOH, only Al₂(SO₄)₃ reacts with it and NaCl does not react with NaOH due to presence of common ion (Na⁺). On reaction gelatinous precipitate of aluminium hydroxide [Al(OH)₃] is produced. The balanced chemical reaction is represented as-
Al₂(SO₄)₃ + 6NaOH → 2Al(OH)₃ + 3Na₂SO₄
On this reaction, 0.495 g = 0.495/78 moles =6.346 X 10⁻³ moles of Al(OH)₃.
As per balanced reaction, two moles of Al(OH)₃ is produced from one mole Al₂(SO₄)₃. So, 6.346 X 10⁻³ moles of Al(OH)₃ is produced from (6.346 X 10⁻³)/2 moles=3.173 X 10⁻³ moles of Al₂(SO₄)₃= 3.173 X 10⁻³ X 342 g of Al₂(SO₄)₃=1.085 g of Al₂(SO₄)₃.
So, mass percentage of Al₂(SO₄)₃ is= (amount of Al₂(SO₄)₃/total amount of mixture)X100 = 
=74.8 %.
