OH- is the ion that increases the concentration of a base
M ( HCl ) = ?
V ( HCl ) = 25.5 mL in liters : 25.5 / 1000 => 0.0255 L
M ( NaOH ) = 0.113 M
V ( NaOH ) = 51.2 mL / 1000 => 0.0512 L
number of moles NaOH:
n = M x V
n = 0.113 x <span> 0.0512 => 0.0057856 moles of NaOH
mole ratio:
</span><span>HCl + NaOH = NaCl + H2O
</span><span>
1 mole HCl -------------- 1 mole NaOH
( moles HCl ) ----------- </span><span> 0.0057856 moles NaOH
</span>
(moles HCl ) = <span> 0.0057856 x 1 / 1
</span>
= <span> 0.0057856 moles of HCl
</span>
M ( HCl ) = n / V
M = 0.0057856 / <span>0.0255
</span>
= 0.227 M
Answer A
hope this helps!
Answer:
Reactant : A combustion of hydrocarbon.
Explanation:
It is known that when hydrocarbons are involved in combustion, they produce carbon dioxide and water.
CxHy + (x+y/4)O2 ===> xCO2 + y/2H2O
Answer:
In the final solution, the concentration of sucrose is 0.126 M
Explanation:
Hi there!
The number of moles of solute in the volume taken from the more concentrated solution will be equal to the number of moles of solute in the diluted solution. Then, the concentration of the first solution can be calculated using the following equation:
Ci · Vi = Cf · Vf
Where:
Ci = concentration of the original solution
Vi = volume of the solution taken to prepare the more diluted solution.
Cf = concentration of the more diluted solution.
Vf = volume of the more diluted solution.
For the first dillution:
26.6 ml · 2.50 M = 50.0 ml · Cf
Cf = 26.6 ml · 2.50 M / 50.0 ml
Cf = 1.33 M
For the second dilution:
16.0 ml · 1.33 M = 45.0 ml · Cf
Cf = 16.0 ml · 1.33 M / 45.0 ml
Cf = 0.473 M
For the third dilution:
20.0 ml · 0.473 M = 75.0 ml · Cf
Cf = 20.0 ml · 0.473 M / 75.0 ml
Cf = 0.126 M
In the final solution, the concentration of sucrose is 0.126 M
