Ask question

Ask question

All categories

2 years ago

15

Sort these isotopes by whether they are most likely to undergo fusion or fission.

1 answer:

Maslowich2 years ago

4 0

Answer:

Uranium-233- fission

Plutonium-239- fission

Plutonium-241- fission

Hydrogen-3 fusion

Hydrogen-1 fusion

Helium-3 fusion

Explanation:

In nuclear fission, heavy nuclear disintegrate into smaller nuclei when bombarded with particles such as neutrons. Fission reaction is common among nuclei having a high atomic number such as plutonium and uranium.

Fusion occurs between two light nuclei such as hydrogen or helium. It involves the combination of two lighter elements to give a heavier element with the release of tremendous amount of energy.

You might be interested in

This graph illustrates that, under conditions of constant pressure, as the temperature of liquid water increases, its volume ___

Lunna [17]

it varies.........................

8 0

2 years ago

If a hot air balloon has an initial volume of 1000 L at 50 ∘C, what is the temperature (in ∘C) of the air inside the balloon if

puteri [66]

<span>(P1/T1) = (P2/T2)

T must be in kelvin first!</span>

3 0

3 years ago

Sam places 5 g of baking soda in 5 g of liquid vinegar, and notices fizzing. After all the fizzing stops, 5 g of liquid remains.

anygoal [31]

Answer: G

5+5=10

10-5=5

5 g of gas escaped

3 0

2 years ago

Read 2 more answers

Aspirin (acetylsalicylic acid, C9H8O4) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolv

Triss [41]

Answer:

1. $3.57\times 10^{-4}$ was the $K_a$ value calculated by the student.

2. $5.93\times 10^{-4}$ was the $K_b$ of ethylamine value calculated by the student.

Explanation:

1.

The $pH$ value of Aspirin solution = 2.62

$pH=-\log[H^+]$

$[H^+]=10^{-2.62}=0.00240 M$

Moles of s asprin = $\frac{2.00 g}{180 g/mol}=0.01111 mol$

Volume of the solution = 0.600 L

The initial concentration of Aspirin = c = $\frac{0.01111 mol}{0.600 L}=0.0185 M$

$HAs\rightleftharpoons As^-+H^+$

initially

c 0 0

At equilibrium

(c-x) x x

The expression of dissociation constant :

$K_a=\frac{[As^-][H^+]}{[HAs]}$ :

$K_a=\frac{x\times x }{(c-x)}$

$=\frac{0.00240 M\times 0.00240 M}{(0.0185-0.00240 )}$

$K_a=3.57\times 10^{-4}$

$3.57\times 10^{-4}$ was the $K_a$ value calculated by the student.

2.

The $pH$ value of ethylamine = 11.87

$pH+pOH=14$

$pOH=14-11.87=2.13$

$pOH=-\log[OH^-]$

$[OH^-]=10^{-2.13}=0.00741 M$

The initial concentration of ethylamine = c = 0.100 M

$C_2H_5NH_2+H_2O\rightleftharpoons C_2H_5NH_3^{+}+OH^-$

initially

c 0 0

At equilibrium

(c-x) x x

The expression of dissociation constant :

$K_b=\frac{[C_2H_5NH_3^{+}][OH^-]}{[C_2H_5NH_2]}$ :

$K_b=\frac{x\times x}{(c-x)}$

$=\frac{0.00741\times 0.00741}{(0.100-0.00741)}$

$K_b=5.93\times 10^{-4}$

$5.93\times 10^{-4}$ was the $K_b$ of ethylamine value calculated by the student.

3 0

3 years ago

kap26 [50]

Answer:

what you've asked so I guess the answer will be 10 days

8 0

2 years ago