Answer 1) When a strong acid like

reacts with

usually the equilibrium shifts to the right because
As per the Le chatelier's principle "if in any reaction, a dynamic equilibrium is disturbed by changing the any of the conditions, the position of equilibrium moves to counteract the change." So, in the given reaction when

reacts with

it generates carbon dioxide and water as a by product, if we are adding

it will remove some of the

molecule from the reaction mixture, which then tends to shift the equilibrium towards right.
Answer 2) The same would be observed in this case, if we replace

with HCl it will shift the equilibrium to the right as their will be generation of AgCl as the precipitate.
As per the definition of Le Chatelier's principle if we add reactants in the reaction the equilibrium will tend to move towards right, also if we replace the products or remove it then too it will shift the equilibrium towards right. So, in this reaction you are removing

and

ions from the solution.
Use Charles' Law: V1/T1 = V2/T2. We assume the pressure and mass of the helium is constant. The units for temperature must be in Kelvin to use this equation (x °C = x + 273.15 K).
We want to solve for the new volume after the temperature is increased from 25 °C (298.15 K) to 55 °C (328.15 K). Since the volume and temperature of a gas at a constant pressure are directly proportional to each other, we should expect the new volume of the balloon to be greater than the initial 45 L.
Rearranging Charles' Law to solve for V2, we get V2 = V1T2/T1.
(45 L)(328.15 K)/(298.15 K) = 49.5 ≈ 50 L (if we're considering sig figs).
There are different formula you need to keep in mind when solving for [OH-]
Given that pH = 6.10
pH + pOH = 14
6.10 + pOH = 14
pOH = 7.9
[OH-] = 10^(-pOH)
[OH-] = 10^(-7.9)
[OH-] = 0.000000013
[OH-] = 1.3 x 10^-8
<h2>
<u>Answer: [OH-] = 1.3 x 10^-8</u></h2>