<span>A compound is found to be 40.0% carbon, 6.7% hydrogen and 53.5% oxygen. Its molecular mass is 60. g/mol.
</span>Q1)
Empirical formula is the simplest ratio of whole numbers of components making up a compound.
the percentages have been given, therefore we can calculate for 100 g of the compound.
C H O
Mass in 100 g 40.0 g 6.7 g 53.5 g
Molar mass 12 g/mol 1 g/mol 16 g/mol
Number of moles 40.0/12= 3.33 6.7/1 = 6.7 53.5/16 = 3.34
Divide by the least number of moles
3.33/3.33 = 1 6.7/3.33 = 2.01 3.34/3.33 = 1.00
after rounding off
C - 1
H - 2
O - 1
Empirical formula - CH₂O
Q2)
Molecular formula is the actual number of components making up the compound.
To find the number of empirical units we have to find the mass of one empirical unit.
Mass of one empirical unit = CH₂O - 12 + (1x2) + 16 = 30 g
Mass of one mole of compound = 60 g
Number of empirical units = 60 g / 30 g = 2
Therefore molecular formula - 2(CH₂O)
Molecular formula - C₂H₄O₂
Answer:
1 litre = 1000 millilitres
Answer is: 5,75·10⁻¹.
Kf = 2,3·10⁶ 1/s.
K = 4,0·10⁸ 1/s.
Kr = ?
Kf - <span>forward rate constant.
K - </span><span>equilibrium constant.
Kr - </span><span>reverse rate constant.
</span>Since both Kf and Kr are constants at a given temperature, their ratio is also a constant that
is equal to the equilibrium constant K.<span>
K = Kf/Kr.
Kr = Kf/K = </span>2,3·10⁶ 1/s ÷ 4,0·10⁸ 1/s = 5,75·10⁻¹.
Answer:
Distillation and using Chromatography
Explanation:
Distillation:
based on using boiling point.
Ex: separating a mixture of water and sugar by boiling the water away.
Chromatography:
based on absorption
Ex: separating the different colours that make up a black marker
Explanation:
This is correct!
Ions that exist in both the reactant and product side of the equation are referred to as spectator ions. Overall, they do not partake in the reaction. If they are present on both sides of the equation, you can cancel them out.
An example is;
Na+(aq) + Cl−(aq) + Ag+(aq) + NO3−(aq) → Na+(aq) + NO3−(aq) + AgCl(s)
The ions; Na+, NO3−(aq) would be cancelled out to give;
Cl−(aq) + Ag+(aq) → AgCl(s)
