Volume of NaOH required to react = 145.5 ml
<h3>Further explanation</h3>
Reaction
CO₂(g)
+ 2 NaOH(aq) ⇒Na₂CO₃(aq) + H₂O(l)
The volume of CO₂ : 0.45 L
mol CO₂ at STP (O C, 1 atm) ⇒ at STP 1 mol gas 22.4 L :
From the equation, the mol ratio of CO₂ : NaOH = 1 : 2, so mol NaOH :
Then volume of NaOH :
The dominant gene become one that had insulin and was dependent. And the ressesive became non insulin because those who had it could survive easier
Make sure that you understand what they are asking you from this question, as it can be confusing, but the solution is quite simple. They are stating that they want you to calculate the final concentration of 6.0M HCl once a dilution has been made from 2.0 mL to 500.0 mL. They have given us three values, the initial concentration, initial volume and the final volume. So, we are able to employ the following equation:
C1V1 = C2V2
(6.0M)(2.0mL) = C2(500.0mL)
Therefore, the final concentration, C2 = 0.024M.
The answer is the first option (a)
