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diamong [38]
4 years ago
13

Which of the following best describes the changes involved in a fission reaction?

Chemistry
1 answer:
never [62]4 years ago
8 0
The correct answer is this: THE NUCLEUS OF AN ATOM SPLITS INTO FRAGMENTS, RELEASING A LARGE AMOUNT OF ENERGY.
Nuclear fission is the process in which the nucleus of a radioactive element split into two different nucleic of smaller sizes of different elements with a large release of energy. Nuclear fission process is usually used to provide energy for electricity generation. 
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Which statement about the scientific veiws is correct?
hammer [34]
I think the answer will be b cuh
4 0
4 years ago
What is the % dissociation of a solution of acetic acid if at equilibrium the solution has a pH = 4.74 and a pKa = 4.74?
Ymorist [56]

Answer:

\% diss = 50\%

Explanation:

Hello there!

In this case, when considering weak acids which have an associated percent dissociation, we first need to set up the ionization reaction and the equilibrium expression:

HA\rightleftharpoons H^++A^-\\\\Ka=\frac{[H^+][A^-]}{[HA]}

Now, by introducing x as the reaction extent which also represents the concentration of both H+ and A-, we have:

Ka=\frac{x^2}{[HA]_0-x} =10^{-4.74}=1.82x10^{-5}

Thus, it is possible to find x given the pH as shown below:

x=10^{-pH}=10^{-4.74}=1.82x10^{-5}M

So that we can calculate the initial concentration of the acid:

\frac{(1.82x10^{-5})^2}{[HA]_0-1.82x10^{-5}} =1.82x10^{-5}\\\\\frac{1.82x10^{-5}}{[HA]_0-1.82x10^{-5}} =1\\\\

[HA]_0=3.64x10^{-5}M

Therefore, the percent dissociation turns out to be:

\% diss=\frac{x}{[HA]_0}*100\% \\\\\% diss=\frac{1.82x10^{-5}M}{3.64x10^{-5}M}*100\% \\\\\% diss = 50\%

Best regards!

6 0
3 years ago
Examine the photo of quartz below in which way does the quartz break
AveGali [126]
Can u add a link please 

4 0
3 years ago
Complete the following single replacement reaction. If they don’t react, just write “NR”
stiv31 [10]

We have to complete all the given reactions.

1. Fe(s) + CuCl₂ → Cu + FeCl₂

2. Cu(s) + FeCl₂(aq)  → NR (no reaction takes place)

3. K(s) + NiBr2(aq) → NR (no reaction takes place)

4. Ni(s) + KBr(aq) → K + NiBr₂

5. Zn(s) + Ca(NO₃)₂(aq) → NR (no reaction)

6. Ca(s) + Zn(NO₃)₂(aq) → Zn(s) + Ca(NO₃)₂(aq)

4 0
3 years ago
A second-order reaction has a rate law: rate = k[a]2, where k = 0.150 m−1s−1. If the initial concentration of a is 0.250 m, what
Cerrena [4.2K]

Rate law for the given 2nd order reaction is:

Rate = k[a]2

Given data:

rate constant k = 0.150 m-1s-1

initial concentration, [a] = 0.250 M

reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s

To determine:

Concentration at time t = 300 s i.e. [a]_{t}

Calculations:

The second order rate equation is:

1/[a]_{t} = kt +1/[a]

substituting for k,t and [a] we get:

1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M

1/[a]t = 49 M-1

[a]t = 1/49 M-1 = 0.0204 M

Hence the concentration of 'a' after t = 5min is 0.020 M



7 0
3 years ago
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