Answer:
C)
Explanation:
The warmer the molecules, the higher the average kinetic energy.
"As stated in the kinetic-molecular theory, the temperature of a substance is related to the average kinetic energy of the particles of that substance. When a substance is heated, some of the absorbed energy is stored within the particles, while some of the energy increases the motion of the particles. This is registered as an increase in the temperature of the substance." -lumen learning
(I know for sure but what think... hope this helped)
Answer:
D.<em> </em>Radiation
<em>Explanation:</em>
The Sun always transfers its energy to any planet (Including Earth) through <u>solar radiation, infrared, and UV rays.</u> These are all examples of <u>radiation.</u>
I hope this helps you! Good luck with your test/assignment! :D
Answer:
pH=11.
Explanation:
Hello!
In this case, since the data is not given, it is possible to use a similar problem like:
"An analytical chemist is titrating 185.0 mL of a 0.7500 M solution of ethylamine(C2HNH2) with a 0.4800 M solution of HNO3.ThepK,of ethylamine is 3.19. Calculate the pH of the base solution after the chemist has added 114.4 mL of the HNO3 solution to it"
Thus, for the reaction:

Tt is possible to compute the remaining moles of ethylamine via the following subtraction:

Thus, the concentration of ethylamine in solution is:
![[ethylamine]=\frac{0.0816mol}{0.1850L+0.1144L}=0.2725M](https://tex.z-dn.net/?f=%5Bethylamine%5D%3D%5Cfrac%7B0.0816mol%7D%7B0.1850L%2B0.1144L%7D%3D0.2725M)
Now, we can also infer that some salt is formed, and has the following concentration:
![[salt]=\frac{0.0549mol}{0.1850L+0.1144L}=0.1834M](https://tex.z-dn.net/?f=%5Bsalt%5D%3D%5Cfrac%7B0.0549mol%7D%7B0.1850L%2B0.1144L%7D%3D0.1834M)
Therefore, we can use the Henderson-Hasselbach equation to compute the resulting pOH first:
![pOH=pKb+log(\frac{[salt]}{[base]} )\\\\pOH=3.19+log(\frac{0.1834M}{0.2725M})\\\\pOH=3.0](https://tex.z-dn.net/?f=pOH%3DpKb%2Blog%28%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bbase%5D%7D%20%29%5C%5C%5C%5CpOH%3D3.19%2Blog%28%5Cfrac%7B0.1834M%7D%7B0.2725M%7D%29%5C%5C%5C%5CpOH%3D3.0)
Finally, the pH turns out to be:

NOTE: keep in mind that if you have different values, you can just change them and follow the very same process here.
Best regards!
Explanation:
the trend in electropositivity in group 1
Li < Na < K < Rb < Cs