I believe it means the imitations on what you did what were you not able to do to the object. I'm not too sure but I believe that's what it means. I hope I was able to help
<>"Atomic particles. Protons and neutrons are heavier than electrons and reside in the nucleus at the center of the atom. Electrons are extremely lightweight and exist in a cloud orbiting the nucleus. The electron cloud has a radius 10,000 times greater than the nucleus."<>
Answer:
This question is incomplete, the remaining part of the question is:
What is the control group, independent variable and dependent variable?
Control group: Plants placed in 80 degree rooms
Independent variable: Change in temperature
Dependent variable: Change in color of leaves
Explanation:
The independent variable in a scientific experiment is the variable that the experimenter controls or manipulates in order to bring about a change in the dependent variable. In this experiment, the variable manipulated by Justin B is the TEMPERATURE CHANGE.
On the other hand, a variable is said to be dependent if it is the variable that responds to a change made to the independent variable or rather it is the outcome. In this experiment, Justin B is trying to see the outcome on the color change in leaves when exposed to a low temperature, hence, COLOR CHANGE IN LEAVES is the dependent variable.
Control group of an experiment is the group that receives no experimental treatment. It is the group the experimenter considers normal and hence is comparing with his experimental group. In this experiment, Justin B believes the leaves change color in a low temperature, hence, he placed some plants in a lower temperature (60 degree) in order to compare them with when the plants are placed in a higher temperature (80 degree). As far as this experiment is concerned, the plants placed in 80 degrees temperature are believed by Justin B not to undergo color change, hence, they are the CONTROL GROUP while the group he placed in 60 degrees temperature are what he is interested in, making them the EXPERIMENTAL GROUP
Answer:
Only
gives spontaneous reaction.
Explanation:
A redox reaction will be spontaneous if standard reduction potential (
) of the reaction is positive. Because it leads to negative standard gibbs free energy change (
), which is a thermodynamic condition for spontaneity of a reaction.

Where
and
represents standard reduction potential of reduction half cell and standard reduction potential of oxidation half cell.
(1) Oxidation:
; 
Reduction:
; 
So, 
Hence this pair will give spontaneous reaction.
(2) Similarly as above, 
Hence this pair will give non-spontaneous reaction.
(3) Similarly as above, 
Hence this pair will give non-spontaneous reaction.
(4) Similarly as above, 
Hence this pair will give non-spontaneous reaction.
Answer:
Δ S = 26.2 J/K
Explanation:
The change in entropy can be calculated from the formula -
Δ S = m Cp ln ( T₂ / T₁ )
Where ,
Δ S = change in entropy
m = mass = 2.00 kg
Cp =specific heat of lead is 130 J / (kg ∙ K) .
T₂ = final temperature 10.0°C + 273 = 283 K
T₁ = initial temperature , 40.0°C + 273 = 313 K
Applying the above formula ,
The change in entropy is calculated as ,
ΔS = m Cp ln ( T₂ / T₁ ) = (2.00 )( 130 ) ln( 283 K / 313 K )
ΔS = 26.2 J/K