How many milliliters of a 1.5 m h2so4 are needed to neutralize 35ml sample of a 1.5 m solution?
1 answer:
Answer:
1) 17.5 mL
Explanation:
Hello,
In this case, the reaction between sulfuric acid and potassium hydroxide is:
In such a way, we notice a 1:2 molar ratio between the acid and the base, therefore, at the equivalence point we have:
And in terms of concentrations and volumes:
Thus, we solve for the volume of acid:
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Answer:
Volume : 1.25 L
Explanation:
We are given here that the volume ( V ) = 1.50 Liters, the initial moles ( held at 25 °C ) = 3.00 mol, and the final moles ( n
) = 3.00 - 0.5 = 2.5 mol. The final mol is calculated given that 0.50 mol of gas are released from the prior 3.00 moles of gas.
Volume ( V ) = 1.50 L,
Initial moles ( n ) = 3.00 mol,
Final Volume ( n ) = 3.00 - 0.5 = 2.5 mol
Applying the combined gas law, we can calculate the final volume ( V ).
PV
/ n
T
= P
V
/ n
T
- we know that the pressure and temperature are constant, and therefore we can apply the following formula,
V / n
= V
/ n
- isolate V
,
V = V
n
/ n
= 1.50 L
2.5 mol / 3.00 mol = ( 1.5
2.5 / 3 ) L = 1.25 L
The volume of the balloon will be 1.25 L.