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jenyasd209 [6]
3 years ago
11

Plsplsplsplsplsplsplsplsplspls help

Chemistry
1 answer:
MA_775_DIABLO [31]3 years ago
6 0

Answer:

this is difficult but simple to answer

Explanation:

all atoms move in 1 direction no more than 2

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Mrs.Jacobs dropped an object from 10 meters she knows it did 50 joules of work how much did it weigh?
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Rank the compounds in each set in order of increasing acid strength.
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Answer:

See explanation

Explanation:

For this question, we have to remember the effect of an atom with high <u>electronegativity</u> as "Br". If the "Br" atom is closer to the carboxylic acid group (COOH) we will have an <u>inductive effect</u>. Due to the electronegativity of Br, the electrons of the C-H bond would be to the Br, then this bond would be <u>weaker</u> and the compound will be more acid (because is easier to produce the hydronium ion H^+).

With this in mind, for A in the last compound, we have <u>2 Br atoms</u> near to the acid carboxylic group, so, we will have a high inductive effect, then the C-H would be weaker and we will have <u>more acidity</u>. Then we will have the compound with only 1 Br atom and finally, the last compound would be the one without Br atoms.

In B, the difference between the molecules is the <u>position</u> of the "Br" atom in the molecule. If the Br atom is closer to the acid group we will have a <u>higher inductive effect</u> and more <u>acidity</u>.

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Iodine-131 is a beta emitter used as a tracer in radio immunoassays in biological systems. It follows first order kinetics. The
allsm [11]

<u>Answer:</u> The amount of Iodine-131 remain after 39 days is 0.278 grams

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The equation used to calculate rate constant from given half life for first order kinetics:

t_{1/2}=\frac{0.693}{k}

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t_{1/2} = half life of the reaction = 8.04 days

Putting values in above equation, we get:

k=\frac{0.693}{8.04days}=0.0862days^{-1}

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,

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[A] = amount left after decay process = ?

Putting values in above equation, we get:

0.0862=\frac{2.303}{39}\log\frac{8.0}{[A]}

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