M= #moles / L
4.35/.75 = 5.6
Answer:
Because milk has higher KE than ice, KE is transferred from the milk to the molecules of ice.
Explanation:
The best statement that expresses the transfer of kinetic energy(K.E) is that kinetic energy is transferred from the milk to the ice.
Kinetic energy is form of energy due to motion of the particles of a medium. In this regard, we are dealing with heat energy.
- Heat energy is dissipated from a body at higher temperature to one at a lower temperature.
- Ice is at a lower temperature which is 0°C
- Heat will be transferred in form of thermal energy from the body at higher temperature to one with a lower temperature.
- This is from the milk to the molecules of ice.
Thank you for posting your question here at brainly. Below are the choices that can be found elsewhere:
12.88 M
<span>0.1278 M </span>
<span>0.2000 M </span>
<span>0.5150 M
</span>
Below is the answer:
<span>5 times diluted (250/50),so 2.575/5=0.515 M
</span>
I hope it helps.
Answer:
The MAD of city 2 is less than the MAD for city 1, which means the average monthly temperature of city 2 vary less than the average monthly temperatures for City 1.
Explanation:
For comparing the mean absolute deviations of both data sets we have to calculate the mean absolute deviation for both data sets first,
So for city 1:
Now to calculate the mean deviations mean will be subtracted from each data value. (Note: The minus sign is ignored as the deviation is the distance of value from the mean and it cannot be negative. For this purpose absolute is used)
The deviations will be added then.
So the mean absolute deviation for city 1 is 24 ..
For city 2:
Now to calculate the mean deviations mean will be subtracted from each data value. (Note: The minus sign is ignored)
The deviations will be added then.
So the MAD for city 2 is 11.33 ..
So,
The MAD of city 2 is less than the MAD for city 1, which means the average monthly temperature of city 2 vary less than the average monthly temperatures for City 1.
Answer:
E° = 0.65 V
Explanation:
Let's consider the following reductions and their respective standard reduction potentials.
Sn⁴⁺(aq) + 2 e⁻ → Sn²⁺(aq) E°red = 0.15 V
Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
The reaction with the highest reduction potential will occur as a reduction while the other will occur as an oxidation. The corresponding half-reactions are:
Reduction (cathode): Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
Oxidation (anode): Sn²⁺(aq) → Sn⁴⁺(aq) + 2 e⁻ E°red = 0.15 V
The overall cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = 0.80 V - 0.15 V = 0.65 V
