Propane
Oxygen
Methane
Ammonia
Correct answer is C
No explanation has been provided for this answer.
11.00
22.00
33.00
44.00
Correct answer is B
56.00cm3 of a gas at S.T.P weighed 0.11g
Molar volume of the gas = 22.4dm3 = 22400cm3
molar mass = (22400 x 0.11)/56
molar mass = 2464/56 = 44
Vapour density = Molar mass/2 = 44/2 = 22
2.78g
5.56g
8.34g
11.12g
Correct answer is A
HCl + Pb(NO3)2 ===> 2HNO3 + PbCl2
2moles of HCl ===> 1mole of PbCl2
n = CV/1000 = (100 x 0.2)/(1000) = 0.02 mol
n (HCl) = 0.02 mol
2 moles of HCl ===> 1 mole of PbCl2
0.02 mole of HCl ===> x mole of PbCl2
2x = 0.02
x = 0.01 mol
n = m/M
m = 0.01 x (207 + (35.5 x 2)) = 0.01 x 278 = 2.78 g note; HCl is used for the calculation instead of lead(II) nitrate because it is the limiting reactant meaning it will be used up first before lead(11) nitrate
Litmus paper
Phenolphthalein
Methyl orange
Universal indicator
None of these
Correct answer is C
No explanation has been provided for this answer.
pass each gas into lime water
pass each gas into water and test with litmus paper
pass each gas into concentrated sulphuric acid
expose each gas to atmospheric air
expose each gas to fumes of hydrogen chloride
Correct answer is A
No explanation has been provided for this answer.