Answer:
When atom undergoes the alpha emission the original atom convert into the atom having mass number less than 4 and atomic number 2 less than the starting atom.
Explanation:
Alpha radiations are emitted as a result of radioactive decay. The atom emit the alpha particles consist of two proton and two neutrons. Which is also called helium nuclei. When atom undergoes the alpha emission the original atom convert into the atom having mass number less than 4 and atomic number 2 less than the starting atom.
Properties of alpha radiation:
Alpha radiations can travel in a short distance.
These radiations can not penetrate into the skin or clothes.
These radiations can be harmful for the human if these are inhaled.
These radiations can be stopped by a piece of paper.
₉₂U²³⁸ → ₉₀Th²³⁴ + ₂He⁴ + energy
Answer:
The solution will not form a precipitate.
Explanation:
The Ksp of PbI₂ is:
PbI₂(s) ⇄ 2I⁻(aq) + Pb²⁺(aq)
Ksp = 1.40x10⁻⁸ = [I⁻]²[Pb²⁺] <em>Concentrations in equilibrium</em>
When 328mL of 0.00345M NaI(aq) is combined with 703mL of 0.00802M Pb(NO₃)₂. Molar concentration of I⁻ and Pb²⁺ are:
[I⁻] = 0.00345M × (328mL / (328mL+703mL) =<em> 1.098x10⁻³M</em>
[Pb²⁺] = 0.00802M × (703mL / (328mL+703mL) =<em> 5.469x10⁻³M</em>
<em />
Q = [I⁻]²[Pb²⁺] <em>Concentrations not necessary in equilibrium</em>
If Q = Ksp, the solution is saturated, Q > Ksp, the solution will form a precipitate, if Q < Ksp, the solution is not saturated.
Replacing:
Q = [1.098x10⁻³M]²[5.469x10⁻³M] = 6.59x10⁻⁹
As Q < Ksp, the solution is not saturated and <em>will not form a precipitate</em>.
Answer:
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Explanation:
<u>1. Balanced chemical equation</u>
- Ba(OH)₂(aq) + 2HNO₃(aq) → Ba(NO₃)₂(aq) + 2H₂O(l)
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<u>2. Determine the number of moles of HNO₃ in the solution</u>
- n = M × V = 0.425M × 0.050 liter = 0.02125 mol HNO₃
<u>3. Use the mole ratio from the balanced chemical equation and the number of moles of HNO₃ to determine the number of moles of Ba(OH)₂.</u>
<u>4. Determine the molar concentration of the solution of Ba(OH)₂</u>
- M = 0.010625mol/0.0368liter
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