Answer:- The hydroxide ion concentration of the solution is 
.
Solution:- The formula used to calculate pOH from hydroxide ion is:
![pOH=-log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E-%5D)
When pOH is given and we are asked to calculate hydroxide ion concentration then we multiply both sides by negative sign and take antilog and what we get on doing this is:
![[OH^-]=10^-^p^O^H](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E-%5Ep%5EO%5EH)
pOH is given as 5.71 and we are asked to calculate hydrogen ion concentration. Let's plug in the given value in the formula:
![[OH^-]=10^-^5^.^7^1](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E-%5E5%5E.%5E7%5E1)
![[OH^-]](https://tex.z-dn.net/?f=%5BOH%5E-%5D)
= 0.00000195 or
So, the hydroxide ion concentration of the solution is .
Answer:
The reaction will shift to the left to produce more reactants.
Explanation:
According to the Le- Chatelier principle,
At equilibrium state when stress is applied to the system, the system will behave in such a way to nullify the stress.
The equilibrium can be disturb,
By changing the concentration
By changing the volume
By changing the pressure
By changing the temperature
Consider the following chemical reaction.
Chemical reaction:
6CO₂ + 6H₂O ⇄ C₆H₁₂O₆ + 6O₂
In this reaction the equilibrium is disturb by increasing the concentration of Product.
When the concentration of product is increased the system will proceed in backward direction in order to regain the equilibrium. Because when product concentration is high it means reaction is not on equilibrium state. As the concentration of O₂ increased the reaction proceed in backward direction to regain the equilibrium state and more reactant is formed.
Are able to travel inside to conduct electricity
Answer:
A) 10 moles of CO and 5 moles O2
Answer:
The third period contains eight elements: sodium, magnesium, aluminium, silicon, phosphorus, sulfur, chlorine, and argon. The first two, sodium and magnesium, are members of the s-block of the periodic table, while the others are members of the p-block.
Explanation:
