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
<span>Neutrons to protons.
Neutrons and protons are tiny particles that are within the nucleus. Neutrons and protons make up the nucleus of the cell and the ratio of neutrons determine the stability of the atomic nuclei. The nucleus will become unstable if the ratio of neutrons to protons are not within the appropriate amount.</span>
Answer:
a. 113 min
Explanation:
Considering the equilibrium:-
2N₂O₅ ⇔ 4NO₂ + O₂
At t = 0 125 kPa
At t = teq 125 - 2x 4x x
Thus, total pressure = 125 - 2x + 4x + x = 125 - 3x
125 - 3x = 176 kPa
x = 17 kPa
Remaining pressure of N₂O₅ = 125 - 2*17 kPa = 91 kPa
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
Given that:
The rate constant, k = 
min⁻¹
Initial concentration = 125 kPa
Final concentration = 91 kPa
Time = ?
Applying in the above equation, we get that:-
