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mojhsa [17]
3 years ago
14

develop a hypothesis to explain why the radiation from the 14C source could be shielded more effectively than the radiation from

the 137Cs source. what do your results suggest about the decay mode of each radioactive isotope?
Chemistry
1 answer:
Hoochie [10]3 years ago
4 0
<span>The results suggests that </span>¹³⁷C  is more radioactive and decays faster than ¹⁴C. Also ¹³⁷C penetrates more than ¹⁴C because it has both beta and gamma particle unlike ¹⁴C which has only a beta particle. Gamma particle passes to any kind of material be it fiber glass or metal unlike a beta particle, it cannot pass through these materials.
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What is the mole ratio of AlCl3 to AlSO4 in the above chemical reaction?
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Rank the members of each set of compounds according to the ionic character of their bonds. Most ionic bonds?a) PCl3 PBr3 PF3 Mos
SCORPION-xisa [38]

Explanation:

More is the electronegativity difference between the combining atoms more polar is the compound. Hence, more ionic it will be in nature.

(a)   Electronegativity value of P = 2.19

Electronegativity value of Cl = 3.16

Electronegativity value of Br = 2.96

Electronegativity value of F = 3.98

Electronegativity difference of a P-Cl bond = 3.16 - 2.19 = 0.97

Electronegativity difference of a P-Br bond = 2.96 - 2.19 = 0.77

Electronegativity difference of a P-F bond = 3.98 - 2.19 = 1.79

Since, a P-F bond has the highest electronegativity difference. Therefore, PF_{3} is the most ionic compound and PBr_{3} is the least ionic compound.

(b)   Electronegativity value of B = 2.04

Electronegativity value of N = 3.04

Electronegativity value of C = 2.55

Electronegativity value of F = 3.98

Electronegativity difference of a B-F bond = 3.98 - 2.04 = 1.94

Electronegativity difference of a N-F bond = 3.04 - 3.98 = 0.94

Electronegativity difference of a C-F bond = 3.98 - 2.55 = 1.43

Since, a B-F bond has the highest electronegativity difference. Therefore, BF_{3} is the most ionic compound and NF_{3} is the least ionic compound.

(c)   Electronegativity value of Se = 2.55

Electronegativity value of Te = 2.1

Electronegativity value of Br = 2.96

Electronegativity value of F = 3.98

Electronegativity difference of a Se-F bond = 3.98 - 2.55 = 1.43

Electronegativity difference of a Te-F bond = 3.98 - 2.1 = 1.88

Electronegativity difference of a Br-F bond = 3.98 - 2.19 = 1.02

Since, a Te-F bond has the highest electronegativity difference. Therefore, TeF_{4} is the most ionic compound and BrF_{3} is the least ionic compound.

6 0
3 years ago
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