Nuclear fusion and nuclear fission are two different types of energy-releasing reactions that occur in the nuclei of an atom.
Here are the major differences between the two:
1. To differentiate the two, fission is the splitting of an atom into two or more smaller atoms while fusion is the conjoining or fusion of two or smaller atoms into larger one.
2. Fission does not normally occur in nature while fusion occurs mostly in heavenly bodies such as the stars.
3.Fission produces highly radioactive particles that can be hazardous to both the living things and its habitat or environment while fusion is "clean energy" and "environmental friendly" meaning there are fewer radioactive particles are produced. But if a fission "trigger" is being used, there will be radioactive particles produced.
Among the two nuclear changes, fission is widely used because this reaction produces heat in nuclear reactor. This heat is used to generate steam which operates the turbines to eventually produce electricity.
Answer:
Inherited traits are passed from parent to offspring by information coded in
the DNA or Deoxyribonucleic Acid
Answer:
The Retention factor (rf) value is = 0.2
Explanation:
- Retention factor (Rf) is factor used substances that could be separated using Chromatography. Retention factor determines how fast the component can move on the chromatogram (stationary phase) after elution. Elution occurs when mobile phase (solvent) moves across the stationary phase when the solute has been spotted on the origin.
- Retention factor (Rf) ranges from value between 0 and 1. The closer the value to 1, the faster it can move upon elution. Rf can be calculated.
- Rf value = distance moved by the solute / distance moved by the solvent
= 0.40cm / 2.00cm
= 0.2
95.6 cal
are needed.
Explanation:
Use the following equation:
q
=
m
c
Δ
T
,
where:
q
is heat energy,
m
is mass,
c
is specific heat capacity, and
Δ
T
is the change in temperature.
Δ
T
=
T
final
−
T
initial
Known
m
=
125 g
c
Pb
=
0.130
J
g
⋅
∘
C
T
initial
=
17.5
∘
C
T
final
=
42.1
∘
C
Δ
T
=
42.1
∘
C
−
17.5
∘
C
=
24.6
∘
C
Unknown
q
Solution
Plug the known values into the equation and solve.
q
=
(
125
g
)
×
(
0.130
J
g
⋅
∘
C
)
×
(
24.6
∘
C
)
=
400. J
(rounded to three significant figures)
Convert Joules to calories
1 J
=
0.2389 cal
to four significant figures.
400
.
J
×
0.2389
cal
1
J
=
95.6 cal
(rounded to three significant figures)
95.6 cal
are needed.
