Question

1. the quantity of electricity that would deposit 2.7 g of aluminium during electrolysis is al = 27.0 g, 1f = 96500 c a. 289500 c. b. 18950 c. c. 28950 c. d. 2895 c. 29. what is the ph of a solution containing 0.001 mol dm⁻³ sodium hydroxide? a. 12 b. 7 c. 11 d. 3 30. a neutral atom of neon with atomic number 10 has the same number of electrons as a. k⁺. b. ca²⁺. c. mg²⁺. d. pb²⁺. 31. which of the following compounds would not decompose on heating? a. calcium trioxocarbonate(iv) b. calcium hydroxide c. calcium chloride d. calcium trioxonitrate(v) 32. which of the following atoms has the highest number of electrons in its valence shell? a. ₁₉k 2 8 8 1 b. ₈o 2 6 c. ₉f 2 7 d. ₇n 2 5 33. the soil around a battery manufacturing factory is likely to contain a high concentration of a a. ca²⁺. b. mg²⁺. c. pb²⁺. d. al³⁺. 34. an example of a heavy chemical is a. calcium chloride. b. ammonia. c. polyvinyl chloride. d. iodine.

Explanation:

28.

Step1: Calculate moles of aluminium

The molar - mass of aluminium ($Al$) is $M = 27.0\ g/mol$. The number of moles of aluminium, $n=\frac{m}{M}$, where $m = 2.7\ g$. So, $n=\frac{2.7\ g}{27.0\ g/mol}=0.1\ mol$.

Step2: Determine the number of moles of electrons

The half - reaction for the deposition of aluminium is $Al^{3 +}+3e^-
ightarrow Al$. For 1 mole of $Al$ deposited, 3 moles of electrons are required. For 0.1 mole of $Al$, the number of moles of electrons, $n_e=0.1\ mol\times3 = 0.3\ mol$.

Step3: Calculate the quantity of electricity

The charge of 1 mole of electrons is 1 Faraday ($F$), and $1F = 96500\ C$. The quantity of electricity $Q=n_e\times F$. Substituting $n_e = 0.3\ mol$ and $F = 96500\ C/mol$, we get $Q=0.3\ mol\times96500\ C/mol = 28950\ C$.

Step1: Determine the $[OH^-]$ concentration

Sodium hydroxide ($NaOH$) is a strong base and dissociates completely in solution as $NaOH
ightarrow Na^++OH^-$. Given $[NaOH]=0.001\ mol/dm^3$, then $[OH^-]=0.001\ mol/dm^3 = 10^{- 3}\ mol/dm^3$.

Step2: Calculate the $pOH$

The formula for $pOH$ is $pOH=-\log[OH^-]$. Substituting $[OH^-]=10^{-3}\ mol/dm^3$, we get $pOH =-\log(10^{-3})=3$.

Step3: Calculate the $pH$

We know that $pH + pOH=14$. So, $pH=14 - pOH$. Substituting $pOH = 3$, we get $pH = 11$.

A neutral neon atom has an atomic number of 10, so it has 10 electrons. $K^+$ has 18 - 1=17 electrons, $Ca^{2 +}$ has 20 - 2 = 18 electrons, $Mg^{2+}$ has 12 - 2=10 electrons, and $Pb^{2+}$ has 82 - 2 = 80 electrons.

Answer:

C. 28950 C

29.