The radioisotope radon-222 has a half-life of 3.8 days. How much of a 73.9-g sample of radon- (3 222 would be left after approxi
1 answer:
Answer:
1.115 g
Explanation:
Applying,
R = R'................... Equation 1
Where R = original sample of radon-222, R' = sample of radon-222 left after decay, n = Total time, t = half-life.
make R' the subject of the equation
R' = R/()............... Equation 2
From the question,
Given: R = 73.9 g, n = 23 days, t = 3.8 days.
Substitute these values into equation 2
R' = 73.9/()
R' = 73.9/
R' = 1.115 g
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Answer:
7.08
Explanation:
To solve this problem we'll use the <em>Henderson-Hasselbach equation</em>:
- pH = pka + log
Where is the ratio of [sodium formate]/[formic acid] and pka is equal to -log(Ka), meaning that:
- pka = -log (1.8x10⁻⁴) = 3.74
We<u> input the data</u>:
- 4.59 = 3.74 + log
And<u> solve for </u>:
- 0.85 = log
=
= 7.08
Explanation:
a. Vanillin(4-hydroxy-3-methoxybenzaldehyde):
In its structure hydroxl group will be present on para position of the benzaldehyde ring and methoxy group on meta position.
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In its structure isopropyl group will be present on ortho position of the phenol ring and methyl group on meta position.
c. Carvacrol (5-isopropyl-2-methylphenol):
In its structure isopropyl group will be present opposite to methyl group which is present ortho position in a phenol ring.
d. Eugenol (4-allyl-2-methoxyphenol):.
In its structure allyl group will be present on para position of the phenol ring and methoxy group on ortho position.
e. Gallic acid (3,4,5-trihydroxybenzoic acid):
In its structure hydroxyl group will be present on both meta positions and on para position of the benzoic acid ring.
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