A because that honestly just makes the most sense
Sodium Hydroxide (NaOH) is also known as lye which is a base (very high ph; Alkaline)
Now, in chemistry, equilibrium is what affects the reaction rate of a reaction. If they are in equilibrium, the concentrations of them will not change (both reactants and products).
Now, lets say that to synthesize a certain chemical, we need it to be in an acidic environment with HCL or some other acid as the catalyst for the reaction.
Well, if we were to add Sodium Hydroxide to this which is very alkaline, the ph would change greatly which affects the reaction rate. If we do not have enough energy to overcome the activation barrier, the reaction will not occur (atleast for a very long time).
However, a common mistake is thinking that a catalyst will affect the equilibrium. This is not true. The reaction will still take place but it will have a very slow reaction rate.
TLDR; Adding a catalyst (like NaOH or Sodium Hydroxide) will not change the equilibrium but instead change the reaction rate. The reaction can still occur, although it can take a very, very long time (like diamonds turning into graphite)
Answer:
Number of moles of solute = 0.6 mole
Mass =13.8 g
Explanation:
Given data:
Number of moles of sodium = ?
Volume = 2.0 L
Molarity = 0.30 M
Mass in gram of sodium= ?
Solution:
<em>Number of moles:</em>
Molarity = number of moles of solute / volume in litter
Number of moles of solute = Molarity × volume in litter
Number of moles of solute = 0.30 M × 2.0 L
Number of moles of solute = 0.6 mole
<em>Mass in gram:</em>
Mass = Number of moles × molar mass
Mass = 0.6 mole× 23 g/mol
Mass =13.8 g
A. I’m not positive but I’m pretty sure
Answer: 55
Explanation: A atom consists of a nucleus and electrons.Nucleus contain neutrons with no charge and protons with positive charge. The electrons bearing negative charge revolve around the nucleus.
An electrically neutral atom contains equal number of protons and electrons. Thus if a neutral element X contains 55 protons, it also contains 55 electrons.
