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Answer:
activation energy hope this is right
The moles of HCl are required to neutralized aqueous solution of the 0.03 KoH
KOH + HCl = KCl + H2O
by use mole ratio between KOH to HCl which is 1:1 the moles Of HCl is also 0.03 moles
Answer:
(a) Moles of ammonium chloride = 0.243 moles
(b) 
(c) 60.68 mL
Explanation:
(a) Mass of ammonium chloride = 13.0 g
Molar mass of ammonium chloride = 53.491 g/mol
The formula for the calculation of moles is shown below:
Thus,
<u>Moles of ammonium chloride = 0.243 moles</u>
(b) Moles of ammonium chloride = 0.243 moles
Volume = 295 mL = 0.295 L ( 1 mL = 0.001 L)

(c) Moles of ammonium chloride = 0.0500 moles
Volume = ?
Molarity = 0.824 M
<u>Volume = 0.05 / 0.824 L = 0.06068 L = 60.68 mL</u>