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bonufazy [111]
10 months ago
7

228 90th then decays by emitting alpha particles until 212 82pb is formed. how many alpha decays will occur during this chain pr

ocess?
Chemistry
1 answer:
andriy [413]10 months ago
8 0

Number of alpha decays is 4.

₉₀Th²³² → ₂He⁴ + ₈₈Ra²²⁸        ∴ α = ₂He⁴

Since, atomic number change by 2 in each alpha decay.

Total charge = 90 -82 = 8

Number of alpha decay = 8/2 = 4.

Alpha decay or α-decay is a type of radioactive decay wherein an atomic nucleus emits an alpha particle (helium nucleus) and thereby transforms or 'decays' into a extraordinary atomic nucleus, with a mass wide variety this is decreased through 4 and an atomic wide variety this is decreased through two.

Alpha decay - A common mode of radioactive decay wherein a nucleus emits an alpha particle (a helium-four nucleus). Beta decay - A commonplace mode of radioactive decay in which a nucleus emits beta particles. The daughter nucleus will have a better atomic number than the original nucleus.

Learn more about alpha decays here:- brainly.com/question/1898040

#SPJ4

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A 0.3146 g sample of a mixture of NaCl ( s ) and KBr ( s ) was dissolved in water. The resulting solution required 45.70 mL of 0
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Answer:

The answer to the question is

The mass percentage of NaCl(s) in the mixture is 49.7%

Explanation:

The given variables are

mass of sample of mixture = 0.3146 g

Volume of AgNO₃ required to react comletely with the chloride ions = 45.70 mL

Concentration of the AgNO₃ added = 0.08765 M

The equations for the reactions oare

NaCl(aq) + AgNO₃ (aq) = AgCl(s) + NaNO₃(aq)

AgNO₃ (aq) + KBr (aq) → AgBr (s) + KNO₃

The equation for the reaction shows one mole of NaCl reacts with one mole of AgNO₃ to form one mole of AgCl

Thus 45.70 mL of 0.08765 M solution of AgNO₃ contains\frac{45.7}{1000} (0.08765) = 0.004 moles

Therefore the sum of the number of moles of Br⁻ and Cl⁻

precipitated out of the solution =  0.004 moles

Thus if the mass of NaCl in the sample = z then the mass of KBr = y

However the mass of the sample is given as 0.3146 g which  means the molarity of the solution is 0.004 moles

given by

\frac{z}{58.44} + \frac{y}{119} = 0.004 moles  and z + y = 0.3146

Therefore z = 0.3146 - y which gives

\frac{(0.3146-y)}{58.44} + \frac{y}{119} = 0.004 moles

-8.7×10⁻³y +0.54×10⁻³ = 0.004

or 8.7×10⁻³y = 1.37769× 10⁻³

y = 0.158 g and z = 0.156 Thus the mass of NaCl = 0.156 g and the mass percentage = 0.156/0.3146×100 = 49.7% NaCl

The mass percentage of NaCl(s) in the mixture is 49.7%

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