The heat required to vaporize 43.9 g of acetone at its boiling point is calculated as below
the heat of vaporization of acetone at its boiling point is 29.1 kj/mole
find the moles of acetone = mass/molar mass
= 43.9g /58 g/mol =0.757 moles
heat (Q) = moles x heat of vaporization
= 29.1 kj/mole x 0.757 moles = 22.03 kj
Adaptation actually and also following control measures on how to avoid it from happening
Answer:
neq N2O4 = 0.9795 mol.....P = 0.5 atm; T = 25°C
Explanation:
ni change eq.
N2O4 1 1 - x 0.8154.....P = 1 atm; T = 25°C
NO2 0 0 + x x
∴ x = neq = Peq.V / R.T.....ideal gas mix
if P = 0.5 atm, T = 25°C; assuming: V = 1 L
⇒ x = neq = ((0.5 atm)(1 L))/((0.082 atm.L/K.mol)(298 K))
⇒ x = neq = 0.0205 mol
⇒ neq N2O4 = 1 - x = 1 - 0.0205 = 0.9795 mol
<span>The symbol for hydronium ion concentration is H+. </span><span>There are quite a few
relationships between [H+] and [OH−]
ions. And because there is a large range of number between 10 to 10</span><span>-15</span><span>
M, the pH is used. pH = -log[H+] and pOH = -log[OH−]. In aqueous solutions, </span><span>[H+
][OH- ] = 10-14. From here we can derive the values of each concentration.</span>
