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Fudgin [204]
3 years ago
6

Please begging you guys someone help me

Chemistry
1 answer:
BaLLatris [955]3 years ago
4 0

Answer:

See Explanation

Explanation:

10) From the options provided for this question, gamma particle is the most energetic. Recall that gamma rays are high energy electromagnetic radiation which are capable of causing a high degree of ionization in matter.

11) The bombardment of U-235 with neutrons leads to the reaction;

U\frac{235}{92}  + n\frac{1}{0}---> I\frac{138}{53}  + Y\frac{95}{39} +3n \frac{1}{0}

Hence

a = 92, b= 95, c= 53

12) In positron emission, a proton is transformed into a neutron. The mass number of the daughter nucleus is the same as its parent but the atomic number decreases by 1.

Hence;

Th\frac{231}{90} -----> e\frac{1}{0} +Ac \frac{231}{89}

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A 0.450 g sample of solid lead(II) nitrate is added to 250 mL of 0.250 M sodium iodide solution. Assume no change in volume of t
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Pb(NO₃)₂ ⇒limiting reactant

moles PbI₂ = 1.36 x 10⁻³

% yield  = 87.72%

<h3>Further explanation</h3>

Given

Reaction(unbalanced)

Pb(NO₃)₂(s) + NaI(aq) → PbI₂(s) + NaNO₃(aq)

Required

  • moles of PbI₂
  • Limiting reactant
  • % yield

Solution

Balanced equation :

Pb(NO₃)₂(s) + 2NaI(aq) → PbI₂(s) + 2NaNO₃(aq)

mol Pb(NO₃)₂ :

= 0.45 : 331 g/mol

= 1.36 x 10⁻³

mol NaI :

= 250 ml x 0.25 M

= 0.0625

Limiting reactant (mol : coefficient)

Pb(NO₃)₂ : 1.36 x 10⁻³ : 1 = 1.36 x 10⁻³

NaI : 0.0625 : 2 = 0.03125

Pb(NO₃)₂ ⇒limiting reactant(smaller ratio)

moles PbI₂ = moles Pb(NO₃)₂ = 1.36 x 10⁻³(mol ratio 1 : 1)

Mass of PbI₂ :

= mol x MW

=  1.36 x 10⁻³ x 461,01 g/mol

= 0.627 g

% yield = 0.55/0.627 x 100% = 87.72%

7 0
3 years ago
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