Math

• Notes
• HW 2.6
• HW 2.7
• HW 3.1
• HW 3.2
• HW 3.3
• HW 3.4
• HW 3.5
• HW 4.1
• HW 4.2
• HW 4.3
• HW 4.4
• HW 4.5
• Exam November 9th
• Exam November 9th Corrections
• Test Review 03-08-2016

• Chem 7.31 - 7.45

7.31 (a)

$Mg(s) + AgNO_3(aq) \longrightarrow Mg(NO_3)(aq) + Ag(s)$
$\color{blue}{Mg(s) + 2AgNO_3(aq) \longrightarrow Mg(NO_3)_2(aq) + 2Ag(s)}$

7.31 (b)

$Al(s) + CuSO_4(aq) \longrightarrow Al_2(SO_4)_3(aq) + Cu(s)$
$\color{blue}{2Al(s) + 3CuSO_4(aq) \longrightarrow 3Cu(s) + Al_2(SO_4)_3(aq)}$

7.31 (c)

$Pb(NO_3)_2(aq) + NaCl(aq) \longrightarrow PbCl_2(s) + NaNO_3(aq)$

$X = NO_3$

$Pb(X)_2 + 2NaCl \longrightarrow PbCl_2 + 2NaX$

$\color{blue}{Pb(NO_3)_2(aq) + 2NaCl(aq) \longrightarrow PbCl_2(s) + 2NaNO_3(aq)}$

7.31 (d)

$Al(s) + HCl(aq) \longrightarrow AlCl_3(aq) + H_2(g)$
$\color{blue}{2Al(s) + 6HCl(aq) \longrightarrow 2AlCl_3(aq) + 3H_2(g)}$

7.32 (a)

$Zn(s) + HNO_3(aq) \longrightarrow Zn(NO_3)_2(aq) + H_2(g)$

$X = NO_3$

$Zn(s) + HX(aq) \longrightarrow Zn(X)_2(aq) + H_2(g)$

$Zn(s) + 2HX(aq) \longrightarrow Zn(X)_2(aq) + H_2(g)$

$\color{blue}{Zn(s) + 2HNO_3(aq) \longrightarrow Zn(NO_3)_2(aq) + H_2(g)}$

7.32 (b)

$Al(s) + H_2SO_4(aq) \longrightarrow Al_2(SO_4)_3(aq) + H_2(g)$

$\color{red}{X} = SO_4$

$Al(s) + H_2\color{red}{X}(aq) \longrightarrow Al_2(\color{red}{X})_3(aq) + H_2(g)$

$2Al(s) + 3H_2\color{red}{X}(aq) \longrightarrow Al_2(\color{red}{X})_3(aq) + 3H_2(g)$

$\color{blue}{2Al(s) + 3H_2SO_4(aq) \longrightarrow Al_2(SO_4)_3(aq) + 3H_2(g)}$

7.32 (c)

$K_2SO_4(aq) + BaCl_2(aq) \longrightarrow KCl(aq) + BaSO_4(s)$

$\color{blue}{K_2SO_4(aq) + BaCl_2(aq) \longrightarrow 2KCl(aq) + BaSO_4(s)}$

7.32 (d)

$CaCO_3(s) \longrightarrow CaO(s) + CO_2(g)$

$\color{blue}{CaCO_3(s) \longrightarrow CaO(s) + CO_2(g)}$

7.33 (a)

$2Al_2O_3(s) \rightarrow  4Al(s) + 3O_2(g)$
= Decomposition (Aluminium oxide reacts to produce aluminium and oxygen)

7.33 (b)

$Br_2(g) + BaI_2(s) \rightarrow BaBr_2(s) + I_2(g)$
= Single replacement (Bromine displaces the iodine in a bromine and barium iodide reaction)

7.33 (c)

$2C_2H_2(g) + 5O_2(g) \rightarrow 4CO_2(g) + 2H_2O(g)$
= Combustion (Hydrocarbon burning oxygen to produce carbon dioxide and water)

7.33 (d)

$BaCl_2(aq) + K_2CO_3(aq) \rightarrow BaCO_3(s) + 2KCl(aq)$
= Double replacement

7.33 (e)

$Pb(s) + O_2(g) \rightarrow PbO_2(s)$
= Combination or synthesis (Lead and oxygen form lead oxide)

7.36 (a)

$CuO + 2HCl \rightarrow CuCl_2 + H_2O$
= Double replacement (Copper oxide and hydrochloric acid form copper chloride and water)

7.36 (b)

$2Al + 3Br_2 \rightarrow 2AlBr_3$
= Combination (Aluminium and bromine for aluminum bromide)

7.36 (c)

$2C_2H_2 + 5O_2 \rightarrow  4CO_2 + 2H_2O$
= Combustion (Acetylene burns oxygen to produce carbon dioxide and water)

7.36 (d)

$Fe_2O_3 + 3C \rightarrow 2Fe + 3CO$
= Single replacement (Reaction between rust and carbon causes the carbon to replace the iron to form iron and carbon oxide)

7.36 (e)

$C_6H_{12}O_6 \rightarrow 2C_2H_6O + 2CO_2$
= Decomposition (A glucose reaction forms ethanol and carbon dioxide)

7.38 (a)

$Ca(s) + O2(g) \rightarrow CaO_2$
(Combination)

7.38 (b)

$C_6H_6(g) + O_2(g) \rightarrow 6CO_2 + 3H_2O$

$C_xH_y + ZO_2(g) \rightarrow XCO_2(g) + \frac{Y}{2}H_2O$

(Combustion)

7.38 (c)

$PbO_2 \rightarrow Pb + O_2$
(Decomposition)

7.38 (d)

$KI + Cl_2 \rightarrow 2KCl + 2I$
(Single replacement. Potassium iodide + Chlorine = Potassium chloride and iodine)

7.38 (e)

$CuCl_2(aq) + Na_2S(aq) \rightarrow CuS + 2NaCl$
(Double replacement)

7.39 (a) Reduction

7.39 (b) Oxidation

7.39 (c) Reduction

7.39 (d) Reduction

7.41 (a) $Zn$ is oxidized, $Cl_2$ is reduced

7.41 (b) $Br$ is oxidized, $Cl_2$ is reduced

7.41 (c) $Pb$ is reduced, $O$ is oxidized

7.41 (d) $Fe$ is reduced, $Sn$ is oxidized

7.42 (a) $Li$ is oxidised, $F$ is reduced

7.42 (b) ???

7.42 (c) $Al$ is oxidised, $Sn$ is reduced

7.42 (d) ???

7.43 (a) $Fe$ gains an electron: reduction

7.43 (b) $Fe$ loses an electron: oxidation

7.44 $\space$ $Cl^-$ - The charge of each $Cl$ atom in $Cl_2$ is $0$. During the reaction, the charge of $Cl$ is reduced from $0$ to $-1$. So, $Cl_2$ is reduced: reduction

7.44

$C_{18}H_{32}O_2 + 2H_2 \longrightarrow C_{18}H_{36}O_2$

Gain of 2 Hydrogen: reduction

BACK TO TOP