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.
Explanation:
Solar energy is the radiant energy emmitted from the Sun. It is the electromagnetic energy
the mass of aluminum oxide (101.96 g/mol) produced from 1.74 g of manganese(iv) oxide (86.94 g/mol) is 1.36g
The reaction is 3 MnO2 + 4 Al ------ 2Al2o3+ Mn
3 mole of manganese oxide give 2 moles of aluminum oxide so by the reaction n( MnO2)/3 =n(al203)2
the formula is n= mass/M so, now substituting values
m (Al2O3)= m(MnO2) X 2 X M (Al2O3) / M(MnO2 X3
so, by substituting values, 2 X101.96 X1.74g / 3 X 86.94 =1.36g
so mass of aluminum oxide obtained = 1.36g
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In order to change celcius to kelvin always add 73 to it leaving you with -195.93
