Answer:
q = -6464.9 kJ
Explanation:
We are given that the heat of combustion is ∆H° = −394 kJ per mol of carbon.Therefore what we need to do is calculate how many moles of C are in the lump of coal by finding its mass since the density is given.
vol = 5.6 cm x 5.1 cm x 4.6 cm = 131.38 cm³
m = d x v = 1.5 g/cm³ x 131.38 cm³ = 197.06 g
mol C = m/MW = 197.06 g/ 12.01g/mol = 16.41 mol
q = −394 kJ /mol C x 16.41 mol C = -6464.9 kJ
Answer: HCI + KOH → KCI + H20
Explanation:
HCI(aq) + KOH(aq) → KCI(aq) + H20(l)
Acid + base → Salt + Water.
The above is a neutralization reaction in which an acid, aqeous HCl reacts completely with an appropriate amount of a base, aqueous KOH to produce salt, aqueous KCl and water, liquid H2O only.
This is a neutralization reaction since, the hydrogen ion, H+, from the HCl is neutralized by the hydroxide ion, OH-, from the KOH to form the water molecule, H2O and salt, KCl only.
Molar mass NaOH = 40.0 g/mol
Volume in liters of solution :
5 mL / 1000 => 0.005 L
number of moles :
4 / 40 => 0.1 moles
M = n / V
M = 0.1 / 0.005
= 20 mol/L or 20 M
hope this helps!
