Live practice

Principle and reasoning. Work from the governing physical relation and apply it to every stated part.

(a) In a stationary fluid, pressure rises downward by ρgh. The feet lie below the brain, so the intervening blood column adds pressure.

(b) Air is compressible. Gravity concentrates most atmospheric mass near the ground, so density falls rapidly with altitude; the upper tens of kilometres contain comparatively little mass.

(c) Pressure at a point is the same for every orientation. The force is directional because it equals the scalar pressure times the chosen surface's normal vector.

A point model is valid when size and rotation do not affect the motion under study.

Principle. Apply the conversion factor once per power of the unit; relative density is the ratio to water density.

Equality of ordered pairs requires equality of corresponding coordinates.

Chlorination forms methyl radicals in the propagation step. A termination collision between two methyl radicals creates a carbon–carbon bond.

Determine hybridisation from the number and type of electron domains around each carbon atom. A saturated carbon with four sigma bonds is (sp^3); a carbon in one double bond is (sp^2); and a carbon in a triple bond or in two cumulative double bonds is (sp).

(a) In (mathrm{CH_2=C=O}), the terminal (mathrm{CH_2}) carbon has one double bond and two C-H sigma bonds, so it is (sp^2). The central carbon forms two cumulative double bonds and is therefore (sp).

(b) In (mathrm{CH_3CH=CH_2}), the methyl carbon has four sigma bonds and is (sp^3). Each carbon of the C=C bond is (sp^2).

(c) In (mathrm{(CH_3)_2CO}), both methyl carbons have four sigma bonds and are (sp^3). The carbonyl carbon contains a C=O double bond and is (sp^2).

(d) In (mathrm{CH_2=CHCN}), the two alkene carbons are (sp^2). The carbon of the (mathrm{Cequiv N}) group is (sp).

(e) In benzene, every carbon forms three sigma bonds and contributes one unhybridised p orbital to the delocalised pi system. Hence every carbon is (sp^2).

Use H=+1, O=−2 (except peroxide), alkali metal=+1 and alkaline-earth metal=+2; set the sum equal to the species charge.

(a) Immediately after expansion, the number of vapour molecules is unchanged but they occupy a larger volume. Their concentration and collision frequency therefore decrease, so the vapour pressure falls.

(b) At that instant the temperature and liquid surface are unchanged, so the rate of evaporation is initially almost unchanged. Fewer vapour molecules strike the liquid surface per second, so the rate of condensation decreases. Thus evaporation temporarily occurs faster than condensation.

(c) More liquid evaporates and the vapour concentration rises. Equilibrium is restored when the rates of evaporation and condensation again become equal. At a fixed temperature, the saturation vapour pressure of a liquid is fixed; therefore, provided some liquid remains, the final vapour pressure returns to its original saturation value.

A state function depends only on the initial and final equilibrium states, regardless of the route followed.

As atoms approach, nucleus-electron attractions lower potential energy until a stable separation is reached. A bond forms when the bonded arrangement has lower energy than the separated atoms.