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Irina-Kira [14]
2 years ago
7

What is the specific activity (in Ci/g) if 1.65 mg of an isotope emits 1.56X10⁶α a particles per second?

Chemistry
1 answer:
marta [7]2 years ago
4 0

The specific activity (in Ci/g) if 1.65 mg of an isotope emits 1.56X10⁶α a particles per second.

<h3>What is isotope?</h3>

Isotopes are two or more types of atoms with the same atomic number (number of protons in their nuclei) and periodic table position (and so belong to the same chemical element), but distinct nucleon numbers due to differing numbers of neutrons in their nuclei. While all isotopes of a given element have nearly identical chemical properties, their atomic weights and physical properties differ.

The term isotope is derived from the Greek roots isos and topos, which means that different isotopes of the same element occupy the same position on the periodic table. Margaret Todd, a Scottish doctor and writer, coined the phrase in 1913 as a recommendation to British chemist Frederick Soddy.

To learn more about isotopes visit:

brainly.com/question/11680817

#SPJ4

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For the equilibrium PCl5(g) PCl3(g) + Cl2(g), Kc = 2.0 × 101 at 240°C. If pure PCl5 is placed in a 1.00-L container and allowed
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Answer:

the equilibrium concentration of [PCl₅] is 3.64*10⁻³ M

Explanation:

for the reaction

PCl₅(g) → PCl₃(g) + Cl₂(g)

where

Kc= [PCl₃]*[Cl₂]/[PCl₅] = 2.0*10¹ M = 20 M

and [A] denote concentrations of A

if initially the mixture is pure PCl₅ , then it will dissociate according to the reaction and since always one mole of PCl₃(g) is generated with one mole of Cl₂(g) , the total number of moles of both at the end is the same → they have the same concentration → [PCl₃(g)] = [Cl₂]=0.27 M

therefore

Kc= [PCl₃]*[Cl₂]/[PCl₅] = 0.27 M* 0.27 M /[PCl₅] = 20 M

[PCl₅]  =  0.27 M* 0.27 M / 20 M = 3.64*10⁻³ M

[PCl₅]  = 3.64*10⁻³ M

the equilibrium concentration of [PCl₅] is 3.64*10⁻³ M

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