A) 2H₂(g) + O₂(g) → 2H₂O(l) + 285.83 kJ
Exothermic
B) 2Mg + O₂ → 2MgO + 1200kJ
Exothermic
First figure out how many grams must freeze and then convert the grams to moles.
<span>Hf = -334 J/g. Convert this to KJ/g by dividing by 1000. (There are 1000 Joules in a kJ). </span>
<span>Hf = -334 J/g ÷ 1000 J/kj = -0.334 kJ/g </span>
<span>Now, divide 100 kJ by -0.334 kJ/g (see how the units are lining up?) </span>
<span>100 kJ ÷ -0.334 kJ/g = 299 g </span>
<span>Now convert this to moles by dividing by the molecular weight of water (18.0g/mole). </span>
<span>299 ÷ 18.0 = 16.6 moles </span>
The amount of carbon in fossils and artifacts decreases when the get older, so you can find out how old an object or fossil is by finding out how much carbon is in it.
NaOH reacts with CH3COOH in 1:1 molar ratio to produce CH3COONa
NaOH + CH3COOH → CH3COONa + H2O
Mol CH3COOH in 52.0mL of 0.35M solution = 52.0/1000*0.35 = 0.0182 mol CH3COOH
Mol NaOH in 19.0mL of 0.40M solution = 19.0/1000*0.40 = 0.0076 mol NaOH
These will react to produce 0.0076 mol CH3COONa and there will be 0.0182 - 0.0076 = 0.0106 mol CH3COOH remaining in solution unreacted . Total volume of solution = 52.0+19.0 = 71mL or 0.071L
Molarity of CH3COOH = 0.0106/0.071 = 0.1493M
CH3COONa = 0.0076 / 0.071 = 0.1070M
pKa acetic acid = - log Ka = -log 1.8*10^-5 = 4.74.
pH using Henderson - Hasselbalch equation:
pH = pKa + log ([salt]/[acid])
pH = 4.74 + log ( 0.1070/0.1493)
pH = 4.74 + log 0.717
pH = 4.74 + (-0.14)
pH = 4.60.
