Balancing Chemical EquationsCu(s)O(g) →CuO(s)28Al+H,SO. H +H0(1) →H(9) +29Si,HO, >SIO,HOFe(s)H,O(g) →H:(g)Fe,O.(s)30NH, +H,OASCI(aq)H;S(aq) →AS S(s) +HCI(aq)co, +31.CuSO. 5H,0(s) →CuSO(s) +H;0(g)32BN +BF, +Fe Os(s)Hg) + Fe(S) +_H;O(1)33.CaSO,+2H,0 +50, →Caso, HSO.CaCO,(s) →CaO(s) +Co.(g)34C,H0,CO,FeS(s)35.C,H,0,0. >CO, +• (s)3HS(04)S(s) >HOH(I) + KS(aq)ZnI, >NalZnKOHaq) →36Na +10.Na(1) +CI,(g)37.HBr0,HBr >H,OBr.11Al(S)H,S0.(aq) + H(g) *Al,(SO,),38AI,C, +CH.NH,OHaq) + HOH(I) (NH.),PO,(aq)(beodHCa(NO,),-3H,0 +LaC,Ca(NO,), La(OH CH3912.13.CHIQ)CO.(9) H,0(1)40.CH,NO, Cl, +CCI,NO, =HCCa,(PO.),SiO,CaSiO,CO P41.14AS)0,9) +(s) o'ivCHEH,0(1)Al,C. +HO +AI(OH4215CHdg)Co(g) *43NafCao +CafNOOH16KINO,KNO,44AICI,>LIAIH.LICI17CocCa +Co,Caf,+SiO,క్రచరి18CH +CO, +HO45HSO.46Casi, +SbCl, >CoCl19.KSOKCIBaSO.TiO,B.C+TiBCO4720KOH +H,50.KSO.48.NH, +NOHO21CofOH)NH.CI -NH.OHCaclz49Sif,NOOHNa SiO,ENeF22So, +CS,CONH.CI +NH,CoCtH050Cao >23.Mg,N,Mg(OH), *NH,Natt51.NaPbCHCI →Pb(CH).24Ca >Cao52Be CBe(OHCH25Na,0,NaOH + 0zHO53Npf,HF>NpF,26Fe O,+H,Fe +H,0NO54.NO,Cuso, +H,0 + 50,+ FONH o'H27Cu +HS0, →

Balancing chemical equations : 1. 2Cu(s) + 1O2(g) → 2CuO 2. 2H2O(l) → 2H2(g) + 1O2(g)

Answered: Cu(s)• Og) CuO(s) DoH Fe(s) HO(g)…

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Chemistry

Cu(s)• Og) CuO(s) DoH Fe(s) HO(g) →H:(9) (60 Fe,O(s) 2. HO(1) → H:(g) 3.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Crystal Structures

Solids are present all around us. They have different properties and applications. Diamond and graphite are two well-known solids. Diamond is the hardest material in the world and is used in making jewelry. Graphite on the other hand is soft. The lead of your pencil is graphite. Why are they so different in appearance and properties despite the fact they are chemically the same? Both are allotropes of carbon. The answer lies in the structure, the way constituent particles are arranged in the solids.

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Balancing Chemical Equations
Cu(s)
O(g) →
CuO(s)
28
Al+
H,SO. H +
H0(1) →
H(9) +
29
Si,H
O, >
SIO,
HO
Fe(s)
H,O(g) →
H:(g)
Fe,O.(s)
30
NH, +
H,O
ASCI(aq)
H;S(aq) →
AS S(s) +
HCI(aq)
co, +
31.
CuSO. 5H,0(s) →
CuSO(s) +
H;0(g)
32
BN +
BF, +
Fe Os(s)
Hg) + Fe(S) +_H;O(1)
33.
CaSO,+2H,0 +
50, →
Caso, HSO.
CaCO,(s) →
CaO(s) +
Co.(g)
34
C,H
0,
CO,
FeS(s)
35.
C,H,0,
0. >
CO, +
• (s)3
HS(04)
S(s) >
HOH(I) + KS(aq)
ZnI, >
Nal
Zn
KOHaq) →
36
Na +
10.
Na(1) +
CI,(g)
37.
HBr0,
HBr >
H,O
Br.
11
Al(S)
H,S0.(aq) + H(g) *
Al,(SO,),
38
AI,C, +
CH.
NH,OHaq) + HOH(I) (NH.),PO,(aq)
(beodH
Ca(NO,),-3H,0 +
LaC,
Ca(NO,), La(OH CH
39
12.
13.
CHIQ)
CO.(9) H,0(1)
40.
CH,NO, Cl, +
CCI,NO, =
HC
Ca,(PO.),
SiO,
CaSiO,
CO P
41.
14
AS)
0,9) +
(s) o'iv
CHE
H,0(1)
Al,C. +
HO +
AI(OH
42
15
CHdg)
Co(g) *
43
Naf
Cao +
Caf
NOOH
16
KINO,
KNO,
44
AICI,>
LIAIH.
LICI
17
Coc
Ca +
Co,
Caf,+
SiO,
క్రచరి
18
CH +
CO, +
HO
45
HSO.
46
Casi, +
SbCl, >
CoCl
19.
KSO
KCI
BaSO.
TiO,
B.C+
TiB
CO
47
20
KOH +
H,50.
KSO.
48.
NH, +
NO
HO
21
CofOH)
NH.CI -
NH.OH
Caclz
49
Sif,
NOOH
Na SiO,
ENeF
22
So, +
CS,
CO
NH.CI +
NH,
CoCt
H0
50
Cao >
23.
Mg,N,
Mg(OH), *
NH,
Natt
51.
NaPb
CHCI →
Pb(CH).
24
Ca >
Cao
52
Be C
Be(OH
CH
25
Na,0,
NaOH + 0z
HO
53
Npf,
HF>
NpF,
26
Fe O,+
H,
Fe +
H,0
NO
54.
NO,
Cuso, +H,0 + 50,
+ FONH o'H
27
Cu +
HS0, →

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