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murzikaleks [220]
3 years ago
5

Hellpppppppppppp Dont Guesss

Chemistry
1 answer:
valina [46]3 years ago
4 0

Answer:

AgF

Agf is the answer.......

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Lithium bromide is a(n) ___________.
bija089 [108]

Answer:

Ionic compound

Explanation:

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It took 55 days for a radioactivity of 1.75 x 1012 Bq to remain 0.135 Ci. What is the half-life of this radioactivity?
Mnenie [13.5K]

From the calculations, the half life of the material is 6.5 days.

<h3>What is radioactivity?</h3>

The term radioactivity has to do with the spontaneous disintegration of a specie.

Uisng the formula;

N=Noe^-kt

N= amount at time t = 0.135 Ci or 4.995 ×10^9 Bq

No = amount initially present =  1.75 x 10^12 Bq

k = rate constant = ?

t = time taken = 55 days

Hence;

4.995 ×10^9  = 1.75 x 10^12e^-55k

4.995 ×10^9/1.75 x 10^12 = e^-55k

2.85 * 10^-3 = e^-55k

ln2.85 * 10^-3 = -55k

k = ln2.85 * 10^-3/-55

k = 0.1066 day-1

Half life = 0.693/ 0.1066 day-1

= 6.5 days

Learn more about radioactivity:brainly.com/question/1770619

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Which is the electron configuration for zinc?
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[AR] 3d10 4s2 is the configuration for zinc

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