Cu(NO3)2>NO2+CuO+O2 balanced: 2Cu(NO3)2=4NO2+2CuO+O2
Answer : The ratio of the protonated to the deprotonated form of the acid is, 100
Explanation : Given,

pH = 6.0
To calculate the ratio of the protonated to the deprotonated form of the acid we are using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[Deprotonated]}{[Protonated]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D)
Now put all the given values in this expression, we get:
![6.0=8.0+\log \frac{[Deprotonated]}{[Protonated]}](https://tex.z-dn.net/?f=6.0%3D8.0%2B%5Clog%20%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D)
As per question, the ratio of the protonated to the deprotonated form of the acid will be:
Therefore, the ratio of the protonated to the deprotonated form of the acid is, 100
Answer:
The difference in temperature causes differences in air pressure between the two spots. ... This causes the jet stream to dip farther south and the winds to blow stronger on either side of the jet stream in winter. The jet stream separates the very cold air mass at higher latitudes from the warmer air mass equatorward.
Explanation:
<u>Answer:</u> The correct statement is low temperature only, because entropy decreases during freezing.
<u>Explanation:</u>
The relationship between Gibb's free energy, enthalpy, entropy and temperature is given by the equation:

Where,
= change in Gibb's free energy
= change in enthalpy
T = temperature
= change in entropy
It is given that freezing of methane is taking place, which means that entropy is decreasing and
is becoming negative. It is also given that the reaction is an exothermic reaction, this means that the
is also negative.
For a reaction to be spontaneous,
must be negative.
![-ve=-ve-[T(-ve)]\\\\-ve=-ve+T](https://tex.z-dn.net/?f=-ve%3D-ve-%5BT%28-ve%29%5D%5C%5C%5C%5C-ve%3D-ve%2BT)
From above equations, it is visible that
will be negative only when the temperature will be low.
Hence, the correct statement is low temperature only, because entropy decreases during freezing.