節樂音育 頁三第張六第日七廿月正年丑乙曆夏
· 1985
中學會考試題預習專欄
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(廿六)
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Solution to Q.37
(8)
(i) The substance is
ammonium chloride,
The equation is NH; (g) + HCl(g) →→NĖ, C1 ( 3 )
(ii) The ring is formed
at a place nearer to Q,
•white ring
(iii)This is because
ammonia diffuses faster than hydrogen chloride (as the. former is lighter
than the latter). As
a consequence,
ammonia meetg
hydrogen chi ride
at a place nearer to 0.
(iv) The electronic
structure of ammonium chloride is:
(v) I would dissociate to form ammonia gas
and hydrogen chloride gas again.
NH, C1(6)-NH; (g) +H¢1 {g} (b)
(1) A is sodium sulphite. (2) B is sulphur dioxide.
(3) NS02+2C1 --
2NaC1+Sg
(4) The reducing property
of 50 as potassium
(ii).
dichromate is a typical oxidizing agent.
(1) When dry nitrogen and
hydrogen gases are passed over a beated catalyst such as iron powder, ammonia gas is obtained. N2+32NH
(2) Hydrogen gas is burnt in oxygen gas to give water vapour.
22+02
2120.
(3) When a mixture of
hydrogen and chlorine gases is exposed to diffused light (or when hydrogen is burnt in chlorine gas) hydrogen is chloride is obtained.
(iii)
2+C12HC1
(1) Ammonia dissolves
readily in water and ionizes to give ammonium ion (NFIL
and hydroxide ion (OH) 0 the
WAH KIU YAT PO
solution is alkaline...
+011
NH3+120
(2) There is no effect. (3) Hydrogen chloride
dissolves readily in water and is completely ionizes to give hydronium jone (0) and
chloride ions (C17).
HCI+H20 30 ci
Solution to 9.38
(i) Nitric acid is rather readily decomposed on exposing to light. Brown bottles can absorb part of the sunlight, thus slowing down the decomposition of nitric acid. Thus, it is usually stored in brown bottles. (ii). Concentrated 50, being an acidic substance, readily Deutralizes ammonia which is an alkaline gas. Thus, the former cannot be used to dry the
latter..
(ii)When a piece of
fresh aluminïum is put into dilute nitric acid, its outer surface is readily oxidized
forming a tough coat
of aluminium oxide, which being. resistant to
the
nitric acid, thus
preventing any
報日僑華
·期星
日八十月三年五八九一年四十七國民華中
maximum mass of H3PO4 obtained from 4.1 g
mol
of P
HA
c4.1
and (KC105) in.
mixturce
1.97-yo
122.5
According to the equation
further attack of
the metal by the acid..
(iv) When lead(Fr) oxide
is added into sulphuric acid or hydrochloric acid, it becomes conted with a layer of insoluble lead(II) sulphate or "... lead(II) chloride respectively, thus stepping any futher reactions.A
Phosphorus is a rather reaction non- me tal and combines readily with oxygen and many other elements, 80, it oçcura naturally as its compounds ratber than in the free... state.
(ii) The other allotrope
"white
(iii).
is called phosphorus).
(1) The formulae, are
PC13 and PC15 respectively.
(2) The oxidation number of phosphours is 13. in PC1, but +5. PCI
(3) Phosphorus(III)
chloride molecule has a 3-D Atructure 38 whown below:
(iv)
(1) Phosphorus is
oxidized by the nitric acid. (2) This is because red
phosphorus is more stable though less reactive at room temperature while whereas phosphorus tursts into flame spontaneously on coming into contact. with air.
(3) It is because
nitrogen dioxide gas produced is harmful to us...
(4) There are three kinde
of salts: sodium dihydrogen phosphate (Nall,PO); disodium
hydrogen phosphate (Na2HPO) trisodium phosphate (NPV).
(5) It is to filter off
any unreacted phosphorus,
(6) M(P)=31 g mol-1
M{H2PO,)=84 y mol According to the
equation... P+SINOPO2+5NO2+H20
1 mol of P gives
1 mol of 11,PU
-10.85 g Solution to Q.39
(i) Magnesium oxide would
be obtained in the boat A and copper powder in the boat B. (ii) Mg+C0g0+CO CO+Cuo→Cu+C02
(iii)The limewater" would
turn milky as carbon dinxide is present.. in the exit gas X.. (iv) Yes; because the
exit gas X may contain unreacted carbon monoxide which in poiṣunous
in nature,
(1) No; because carbon. monoxide cannot. reduce zine(II). oxide.
(2) Yes; lead metal would
be obtained. Pb0+Co→→Pb+C02
(b)
(1) KC10, 2C1+302 (11) Ag*+01→→→→AgC1 (iii)N(KC10 ̧)=122.5.8
..: moi
M(KC)-74.5 g moll M(AgCl)-145.5g mo17
(KC1) in mixture«yg (KC10) in mixture. (1.97-y) £
(KC) in mixture
Now,
2 mol
2mol Imol
1 wol
n(KC) formed on de composing KC103 n(KCI) totally present
n(AgCl) formed
2.87 14351
-0.02 mol
According to the equation
·Ag+C1′′ →AgCl(s):
Amplimol
(KC1): reacted #n(C1") reacted -nagcl) obtained -0.02 mol.......(2) Equation (1) and (2),
we have
1.97-2.0.2
745+122.5
Om solving the equation, y=0.745
by mass of KC, in the original mixture
0.745 1.97 =37.82
x100
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