A solution has an initial concentration of 1.00 mol dm-3. (i) What is its concentration if it is diluted as 1 in 4 parts? (ii) Why should freshly prepared NaOH solutions be used immediately? (iii) Exp

"Diluted as 1 in 4 parts" means that 1 part of the original solution is mixed with 3 parts of solvent to make a total of 4 parts. This implies the final volume is 4 times the initial volume, or the concentration is divided by 4.

Initial concentration ($C_1$) = $1.00 \text{ mol dm}^{-3}$
Dilution factor = 4
Final concentration ($C_2$) = $\frac{C_1}{\text{Dilution factor}}$

The final concentration is $\boxed{0.25 \text{ mol dm}^{-3}}$.

Given:
Number of monomer units ($n$) = 2,000
Relative molecular mass of the polymer ($M_r(\text{polymer})$) = 84,000

The general formula for an alkene is $C_xH_{2x}$.
The relative molecular mass of $C_xH_{2x}$ is $(12 \times x) + (1 \times 2x) = 12x + 2x = 14x$.
Set the calculated monomer mass equal to $14x$:

So, the molecular formula of the alkene is $C_3H_{2 \times 3} = C_3H_6$.

The empirical formula represents the simplest whole-number ratio of atoms in a compound. For $C_3H_6$, the ratio of carbon to hydrogen atoms is $3:6$, which simplifies to $1:2$.
The empirical formula is $CH_2$.
The empirical formula of the alkene is $\boxed{CH_2}$.