Answer:
by principal quantum number (n) and azimuthal quantum number (l)
Explanation:
I used the web to answer so I'm not sure if this is right
250 kJ of energy are removed from a 4.00 x 102 g sample of water at 60˚C. Will the sample of water completely freeze: Yes, because there is enough energy.
<h3>At what temperature would a sample of water freeze?</h3>
- Note from the Facilitator: At certain temperatures, water changes its condition due to temperature variations. At sea level, fresh water changes from a solid to a liquid at 32°F (0°C). Liquid water freezes at temperatures below 32°F (0°C); this temperature is known as the freezing point of water.
- The fact that a single water molecule cannot transform into a solid, liquid, or gas is the answer. These names refer to collective behaviors of water molecules rather than to individual molecules.
- For instance, the solid (ice) has a collection of molecules that are bound together and arranged in a predictable manner. That cannot be accomplished by a single molecule alone
250 kJ of energy are removed from a 4.00 x 102 g sample of water at 60˚C. Will the sample of water completely freeze: Yes, because there is enough energy.
To learn more about water freezing, refer to:
brainly.com/question/15209660
#SPJ9
A thing or substance used for a insulator, for example a substance that does not readily allow the passage of heat or sound
Answer: 
Explanation: For the given reaction:
![Kc=\frac{[Zn^+^2]}{[Ag^+]^2}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BZn%5E%2B%5E2%5D%7D%7B%5BAg%5E%2B%5D%5E2%7D)
Concentrations of the ions are not given so we need to think about another way to calculate Kc.
We can calculate the free energy change using the standard cell potential as:
can be calculated using standard reduction potentials.
Standard reduction potential for zinc is -0.76 V and for silver, it is +0.78 V.
= 
Reduction takes place at anode and oxidation at cathode. As silver is reduced, it is cathode. Zinc is oxidized and so it is anode.
= 0.78 V - (-0.76 V)
= 0.78 V + 0.76 V
= 1.54 V
Value of n is two as two moles of electrons are transferred in the cell reaction F is Faraday constant and its value is 96485 C/mol of electron .
= -297173.8 J
Now we can calculate Kc using the formula:
T = 25+273 = 298 K
R = 
--297173.8 = -(8.314*298)lnKc
297173.8 = 2477.572*lnKc
lnKc = 119.946
