Answer:
A . 2 O₃(g) + 2 NO ⇒ 2 O₂ (g) + 2 NO₂(g)
B . Yes
C. O and NO₃
Explanation:
A. The overall reaction is obtained by adding the individual steps in the reaction mechanism where we will get the reactants and product and the intermediates will cancel.
Thus, adding 1+ 2 +3 we get
2 O₃(g) + 2 NO ⇒ 2 O₂ (g) + 2 NO₂(g)
B. The reaction intermediates are those that are produced from the initial and/or subsequent steps and are consumed later on in the reaction mechanism, but are neither reactants nor products, they just participate.
From this definition it follows that O(g) and NO₃ are reaction intermediates.
C. O and NO₃
Answer:
The water would be neutral, (usually 7). The salt water would be the same (7) and the vinegar would be very acidic. (probably 2).
Explanation:
In general, The more valence electrons a metal has, the stronger its metallic bonds will be because Boron is a metalloid and is ionically bonded.it is too electronegative to release its valence electrons for metallic bonding.As a result, their valence electrons feel a stronger pull from the nucleus (a greater effective nuclear charge) and are less easily released for metallic bonding.
Answer:

Explanation:
Firstly, write the expression for the equilibrium constant of this reaction:
![K_{eq} = \frac{[ADP][Pi]}{ATP}](https://tex.z-dn.net/?f=K_%7Beq%7D%20%3D%20%5Cfrac%7B%5BADP%5D%5BPi%5D%7D%7BATP%7D)
Secondly, we may relate the change in Gibbs free energy to the equilibrium constant using the equation below:

From here, rearrange the equation to solve for K:

Now we know from the initial equation that:
![K_{eq} = \frac{[ADP][Pi]}{ATP}](https://tex.z-dn.net/?f=K_%7Beq%7D%20%3D%20%5Cfrac%7B%5BADP%5D%5BPi%5D%7D%7BATP%7D)
Let's express the ratio of ADP to ATP:
![\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BADP%5D%7D%7B%5BATP%5D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7BK_%7Beq%7D%7D)
Substitute the expression for K:
![\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BADP%5D%7D%7B%5BATP%5D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7BK_%7Beq%7D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7Be%5E%7B-%5Cfrac%7B%5CDelta%20G%5Eo%7D%7BRT%7D%7D%7D)
Now we may use the values given to solve:
![\frac{[ADP]}{[ATP]} = \frac{[Pi]}{K_{eq}} = \frac{[Pi]}{e^{-\frac{\Delta G^o}{RT}}} = [Pi]e^{\frac{\Delta G^o}{RT}} = 1.0 M\cdot e^{\frac{-30 kJ/mol}{2.5 kJ/mol}} = 6.14\cdot 10^{-6}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BADP%5D%7D%7B%5BATP%5D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7BK_%7Beq%7D%7D%20%3D%20%5Cfrac%7B%5BPi%5D%7D%7Be%5E%7B-%5Cfrac%7B%5CDelta%20G%5Eo%7D%7BRT%7D%7D%7D%20%3D%20%5BPi%5De%5E%7B%5Cfrac%7B%5CDelta%20G%5Eo%7D%7BRT%7D%7D%20%3D%201.0%20M%5Ccdot%20e%5E%7B%5Cfrac%7B-30%20kJ%2Fmol%7D%7B2.5%20kJ%2Fmol%7D%7D%20%3D%206.14%5Ccdot%2010%5E%7B-6%7D)