Ask question

Ask question

All categories

3 years ago

14

What occurs during one half-life?

2 answers:

lorasvet [3.4K]3 years ago

8 0

The half-life of a radioactive compound is the time taken for that said isotope to decay or disintegrate so that only half of the initial atoms remain in that compound. During the decay process, the isotope will give off energy and matter, and the way to depict this is indicated by t 1/2.

son4ous [18]3 years ago

6 0

It's B) <span>Half of a parent isotope undergoes radioactive decay to form a daughter isotope.</span>

You might be interested in

Which type of electromagnetic wave has the most energy?

Leto [7]

Answer:

Gamma rays

Explanation:

6 0

2 years ago

Read 2 more answers

Draw the structure of the organic product(s) of the grignard reaction between phosgene (clcocl) and excess phenylmagnesium bromi

Jobisdone [24]

Phosgene on reacting with <span>phenylmagnesium bromide generates benzoyl chloride.

Since, </span>phenylmagnesium bromide is added in excess. It would further react with benzoyl chloride to form benzophenone.

Benzophenone on further reacting with phenylmagnesium bromide, and aqueous treatment, gives triphenylmethanol.

Entire reaction pathways is shown below:

8 0

3 years ago

Read 2 more answers

Identify the spectator ions in this reaction. Check all that apply.

balandron [24]

CN, LI are the only two

4 0

3 years ago

Read 2 more answers

someone pls help me with my chemistry test plsss my teacher changes the questions so I can't search them up. its 21 questions so

shusha [124]

Answer:

Correct option is

B

5 liters of CH

4

(g)NO

2

at STP

No. of molecules=

22.4

5

mol=

22.4

5

×N

A

molecules

A) 5ℊ of H

2

(g)

No. of moles=

2

5

mol=

2

5

×N

A

molecules

B) 5l of CH

4

(g)

No. of moles of CH

4

=

22.4

5

mol=

22.4

5

N

A

molecules

C) 5 mol of O

2

=5N

A

O

2

molecules

D) 5×10

23

molecules of CO

2

(g)

Molecules of 5l NO

2

(g) at STP=5l of CH

4

(g) molecules at STP

Therefore, option B is correct.

7 0

2 years ago

ubstance A undergoes a first order reaction A®B with a half-life, t½, of 20 min at 25 °C. If the initial concentration of A in a

Stells [14]

Answer : The concentration of A after 80 min is, 0.100 M

Explanation :

Half-life = 20 min

First we have to calculate the rate constant, we use the formula :

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

$k=\frac{0.693}{20\text{ min}}$

$k=3.465\times 10^{-2}\text{ min}^{-1}$

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $3.465\times 10^{-2}\text{ min}^{-1}$

t = time passed by the sample = 80 min

a = initial amount of the reactant = 1.6 M

a - x = amount left after decay process = ?

Now put all the given values in above equation, we get

$80=\frac{2.303}{3.465\times 10^{-2}}\log\frac{1.6}{a-x}$

$a-x=0.100M$

Therefore, the concentration of A after 80 min is, 0.100 M

3 0

2 years ago