Answer:
It would react better than what with dilute and will furnish more hydrogen gas.
The mass of He gas = 21.428 g
<h3>Further explanation</h3>
Given
120 liters of He gas
Required
the mass in grams
Solution
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters/mol.
Mol for 120 L :

Mass of He gas(MW=4 g/mol) :

Answer:
0.355 N of HF
Explanation:
The titration reaction of HF with KOH is:
HF + KOH → H₂O + KF
<em>Where 1 mole of HF reacts per mole of KOH</em>
<em />
Moles of KOH are:
0.0296L × (0.120 equivalents / L) = 3.552x10⁻³ equivalents of KOH = equivalents of HF.
As volume of the titrated solution was 10.0mL, normality of HF solution is:
3.552x10⁻³ equivalents of HF / 0.010L =<em> 0.355 N of HF</em>