Chemistry help

Methyl orange is an acidic indicator (red below pH 3 and yellow/ orange above pH 4). Bromothymol blue is a neutral indicator (yellow below pH 6 and blue above pH 8). Phenolphthalein is a basic indicator (colourless below pH 8 and pink/ red above pH 10).

Stomach acid would be red in methyl orange and lemon juice & soda water would be colourless in phenolphthalein. Seawater however, at a pH of 8, would be yellow in methyl orange.

The equation for the Haber process is N₂ + 3H₂ → 2NH₃

This reaction is a synthesis reaction as two chemical species join to form one. The oxidation numbers of N and H also change from 0 to -3 and + 1 respectively, indicating that reduction and oxidation of species have occurred. The reaction is therefore a redox reaction.

Primary alkanols are dehydrated by concentrated sulfuric acid and produce alkenes.

Alkenes readily undergo addition reactions and rapidly decolourise bromine water.

The other alternatives do not produce a product which reacts with bromine water.

C₆H₁₄ is an alkane, C₆H₁₂ could be an alkene or a cycloalkane and C₅H₁₁COOH is an alkanoic acid.

The change in oxidation numbers of

MnO₄^- to Mn²⁺ is from +7 to +2

MnO₂ to Mn(OH)₃ is from +4 to +3

PbO₂ to PbSO₄ is from +4 to +2

VO₂⁺ to VO²⁺ is from +5 to +4

The greatest change occurs in the first example.

To act as an anode (oxidation) the metal must be more reactive than Zn. This eliminates copper and iron.

The lowest theoretical potential is produced from manganese, Mn which is much closer to zinc, Zn on the standard reduction table than calcium, Ca.

Since all molecules have the same molecular weight dispersion forces are roughly equal.

Compounds X and Y however, contain the hydroxyl or -O-H group which produces strong hydrogen bonds between molecules raising the boiling points.

Compound Z does not produce any hydrogen bonds and must a boiling point lower than compound X of 97°C which has hydrogen bonds.

The answer therefore must be A, 31°C

The equation for the fermentation of glucose is C₆H₁₂O₆(aq) → 2CO₂(g) + 2C₂H₅OH(aq)

Step 1. Calculate the number of moles of CO₂

n(CO₂) = m/M, where n(CO₂) = moles of CO₂, m = mass of CO₂ and M = molar mass of CO₂

n(CO₂) = 5.68/44.01 mol

n(CO₂) = 0.120617 ... leave in memory of calculator

Step 2. Determine the number of moles of ethanol

From the fermentation equation above, n(ethanol) = n(CO₂)

n(ethanol) = 0.120617 .... leave in memory of calculator

Step 3. Determine the mass of ethanol

m(ethanol) = n(ethanol) x M(ethanol), where n(ethanol) = moles of ethanol, m = mass of ethanol and M = molar mass of ethanol

m(ethanol) = 0.120617 x 46.068

m(ethanol) = 5.95 g (3 significant figures)

A co-ordinate covalent bond involves one atom providing both electrons.

H₃O⁺, NH₄⁺ and O₃ all contain co-ordinate covalent bonds.

In CO₂, carbon and oxygen atoms each supply a single electron to be shared in the covalent bond and do not produce a co-ordinate covalent bond.

ΔH = m.c.ΔT, where ΔH = energy in J, m = mass of water in kg, c = specific heat constant of water and ΔT = temperature change

21.2 x 10³ = (336.1-215.6)/1000 x 4.18 x 10³ x ΔT,

ΔT = 42.1

Initial temperature = 71.0 - 42.1°C

Initial temperature = 28.9°C which is round up to 29°C

Removing H⁺ ions will move equilibrium to the left and produce more CrO₄^2- ions .

Sodium nitrate and sodium chloride are neutral salts. Ammonium chloride is an acidic salt and sodium acetate a basic salt.

Sodium acetate dissolves in water to produce acetate ions which react with water as follows,

CH₃COO^-(aq) + H₂O(l) <==> CH₃COOH + OH^-(aq).

The hydroxide ions will react with hydrogen ions, shifting equilibrium to the left.

H⁺(aq) + OH^-(aq) → H₂O(l)

The equation for the reaction is 2NaN₃(s) → 2Na(s) + 3N₂(g)

Step 1. Determine the number of moles of N₂

n(N₂) = V/V_M, where n(N₂) = moles of N₂(g), V = volume of gas and V_M of gas at 0°C and 100 kPa pressure.

n(N₂) = 40/ 22.71

n(N₂) = 1.76133... leave in memory of calculator

Step 2. Determine the number of moles of sodium azide, NaN₃

n(NaN₃) = n(N₂) x 2 / 3

n(NaN₃) = 1.1742 ... leave in memory of calculator

Step 3. Determine the mass of sodium azide, NaN₃

m(NaN₃) = n(NaN₃) x M(NaN₃), where m(NaN₃) = mass of sodium azide, n(NaN₃) = moles of sodium azide, M(NaN₃) = Molar mass of sodium azide

m(NaN₃) = 1.1742 ... x 65.02

m(NaN₃) = 76 g rounded off to two significant figures

Step 1. Determine the concentration of copper ions in the unknown solution.

Cu(standard) at 10 ppm, absorbance of 0.4000 : Cu(unknown), absorbance of 0.5000

By ratios 10/ 0.40 = Cu(unknown)/0.5

Therefore Cu(unknown) = 12.5 ppm or 12.5 mg/ L

c(Cu unknown) = m(Cu unknown)/M(Cu unknown) mol. L^-1, where c = concentration, m = mass and M = molar mass

c(Cu unknown) = 12.5 x 10^-3/ 63.55 mol. L^-1

c(Cu unknown) = 0.000196695... leave in memory of calculator.

c(Cu unknown) = 0.000196695 mol. L^-1

Step 2. Determine the number of moles of Cu ions in 100mL of the second solution

n(Cu ions) = c x v, where c = concentration and v = volume.

n(Cu ions) = 0.000196695... x 100/1000

n(Cu ions) = 0.0000196695... mol

Step 3. Write an equation for the reaction and find the precipitate. (Assuming copper(II) which is more common than copper(I))

Cu²⁺(aq) + CO₃^2-(aq) → CuCO₃(s)

The precipitate is CuCO₃(s)

Step 4. Determine the mass of the precipitate

m(CuCO₃) = n(CuCO₃) x M(CuCO₃)

m(CuCO₃) = 0.0000196695... x 123.56

m(CuCO₃) = 2.43 x 10^-3g