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professor190 [17]
3 years ago
9

223Ra decays by alpha emission with a half-life of 11.43 days. For a 1.0g sample of 223Ra, after one half-life, what mass of 223

Ra remains? What has happened to the remaining mass?
Chemistry
1 answer:
Ray Of Light [21]3 years ago
6 0

Explanation:

The reaction that shows the alpha decay of ^{223}_{88}Ra is:

^{223}_{88}Ra\rightarrow ^{219}_{86}Rn+^{4}_{2}He

Half life is the time at which the concentration of the reactant reduced to half. So,

Mass\ of\ ^{223}_{88}Ra\ remained=\frac {Initial\ concentration}{2}=\frac {1.0\ g}{2}=0.5\ g

The remaining mass is converted to ^{219}_{86}Rn.

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Answer:

0.07975 M NaOH

Explanation:

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0.001555 mol KHP / 0.040 L = 0.03888 M KHP

The reaction between KHP and NaOH is one-to-one, so you can just use M1V1 = M2V2 to solve for M2 (the molarity of NaOH).

M1 = 0.03888 M KHP

V1 = 40 mL

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6 Fe2+ (aq) + Cr2O72− (aq) + 14 H+ (aq) → 6 Fe3+ (aq) + 2 Cr3+ (aq) + 7 H2O (aq) If the titration of 23 mL of an iron(II) soluti
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Answer:

1.047 M

Explanation:

The given reaction:

6Fe^{2+}_{(aq)}+Cr_2O_7^{2-}_{(aq)}+14H^+_{(aq)}\rightarrow 6Fe^{3+}_{(aq)}+2Cr^{3+}_{(aq)}+7H_2O_{(aq)}

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The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 15.8 ×10⁻³ L

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Moles=0.254 \times {15.8\times 10^{-3}}\ moles

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Thus,

0.0040132 moles of dichromate react with 6 × 0.0040132 moles of iron(II) solution

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