Question

scrieți 5 oxigizi, 5baze, 5 săruri și calculați raportul atomic și de masa și compoziția procentuala
dau coroana va rog ajutați mă ​

Answer 1

Răspuns:

OXIZI

CaO oxid de calciu

r.at.Ca:O=1:1

r.m.Ca:O=40:16=5:2

comp. proc.

MCaO=40+16=56g/mol

56gCaO.............40gCa.................16gO

100gCaO............xgCa....................ygO

x=100×40/56=calculeaza

y=100×16/56= calculeaza

CO monoxid de carbon

r. at.C:O=1:1

r.m.C:O=12:16=3:4

comp. proc.

MCO=12+16=28g/nol

28gCO.................12gC.....................16gO

100CO....................xgC........................ygO

x=100×12/28=calculeaza

y=100×16/28= calculeaza

Fe2O3 oxid feric

r.a.=2Fe : 3 O= 2: 3

r. m.= Fe : O =112 : 48 =7 : 3

comp. proc.  

MFe2O3=112+ 48=160g/moli

160 gFe2O3.....................112 gFe..............-48 g O

100gFe2O3.........................xFe..................ygO

x=100×112/160= calculeaza

y=100×48 / 160=calculeaza

CO2 bioxid de carbon

RAPORT ATOMIC =C:O=1:2

RAPORT DE MASA=C:O=12:32=3:8

COMPOZITIE PROCENTUALA

MCO2=AC+2×AO=12+2×16=44g

44gCO2.......12gC.......32gO

100gCO2......xgC.........ygO

x=100×12/44=calculează

y=100×32/44=calculează

Al2O3 oxide de aluminiu

r.a.=Al:O= 2 : 3

r.m.= Al:O =54 : 48 =9 : 8

M= 54 + 48= 102 g/moli

comp. proc.  

102g Al2O3......................54g Al..................48g O

100gAl2O3...........................xgAl..................ygO

x= 100× 54 / 102=calculeaza

y=100×48 / 102=calculeaza

BAZE

Mg(OH)2 hidroxid de magneziu

r.a.Mg:O= 1 : 2 : 2

r.m.Mg:O= 24 : 32 : 2 =12 : 16 : 1

comp.proc.

MMg(OH)2= 24 + 32 +2= 58g/moli

58g Mg(OH)2.................24g Mg.................32g O.................2g H

100gMg(OH)2.....................XgMg....................YgO.....................ZgH

X= 100 × 24 / 58=calculeaza

Y=100 ×32 /58=calculeaza

Z= 100 ×2 / 58=calculeaza

Ca(OH)2 hidroxid de calciu

r.at.Ca:O:H=1:2:2

r.m Ca:O:H=40:32:2=20:16:1

Comp. procen.

MCa(OH)2=40+2×16+2×1=74g/molCa(OH)2

74g Ca(OH)2.................40g Ca.................32g O.................2g H

100gCa(OH)2.....................XgCa....................YgO.....................ZgH

X= 100 × 40 / 74=calculeaza

Y=100 ×32 /74=calculeaza

Z= 100 ×2 / 74=calculeaza

NaOH - hidroxid de sodiu / sodă caustică

r.at.  Na : O : H = 1 : 1 : 1

r.m.  Na : O : H  = 23 : 16 : 1

Comp. proc.:

MNaOH=23+16+1=40g/mol

40gNaOH................23gNa..................16gO..................1gH

100gNaOH..................xgNa......................ygO................zgH

x= 100× 23 / 40=calculeaza

y= 100× 16 / 40=calculeaza

z= 100× 1/ 40=calculeaza

Fe(OH)3 hidroxide de fier III

r. at. Fe:O:H= 1:3:3

r. m. Fe:O:H=56:48:3

comp. proc.

MFe(OH)3=56+3×16+3×1==107g/mol

107gNaOH...............56gFe..................3×16gO..................3×1gH

100gNaOH..................xgFe......................ygO......................zgH

x= 100× 56 / 107=calculeaza

y= 100× 16×3 / 107=calculeaza

z= 100× 1×3/ 107=calculeaza

KOH - hidroxid de potasiu

r.at.  K : O : H = 1 : 1 : 1

r.m.  K : O : H  = 39 : 16 : 1

Comp. proc.:

MKOH=39+16+1=56g/mol

56gKaOH................39gK..................16gO..................1gH

100gKOH..................xgK......................ygO................zgH

x= 100× 39 / 56=calculeaza

y= 100× 16 / 56=calculeaza

z= 100× 1/ 56=calculeaza

SARURI

NH₄Cl - clorura de amoniu

r.at :=N : H : Cl = 1 : 4 : 1

r.m. N : H : Cl = 14 : 4 : 35,5

compozitia procentuala

NH₄Cl=14+4×1+35,5=53,5g/mol

53,5 g NH₄Cl .......14 g N ........4 g H ..........35,5 g Cl

100 g  NH₄Cl ........x g N .........y g H ............z g Cl

x=100×14 / 53,5 = calculeaza

y=100×4 / 53,5 =calculeaza

z=100×35,5 /53,5 =calculeaza

Ca(NO3)2 azotat de calcciu

r.at.Ca:N:O = 1:2:6

r.m. Ca:N:O = 40:28:96 = 10:7:24

compozitia procentuala

MCa(NO3)2 = 40+2×14+6×16=164g/mol

164gCa(NO3)2 .......40 g Ca ........2×14 g N ..........6×16gO

100gCa(NO3)2........x g Ca .........y g N ............z g O

x=100×40 / 164 = calculeaza

y=100×2×14 / 164 =calculeaza

z=100×6×16 / 164 =calculeaza

Na2CO3 carbonat de sodiu

r.at. : Na:C:O = 2:1:3

r.m: Na:C:O = 46:12:48 = 23:6:24

compozitia procentuala

MNa2CO3=2×23+12+3×16=106g/mol

106gNa₂CO₃...46gNa...12gC....48gO₂

100gNa₂CO₃....xgNa.......ygC........zgO₂

x=100×46/106=calculeaza

y=100×12/106=calculeaza

z=100×48/106=calculeaza

Ca3(PO4)2fosfat de calciu

r.at. Ca:P:O = 3:2:8.

r.mCa:P:O=120:62:128 = 60:31:64

comp.proc.

MCa3(PO4)2=3×40+2×31+8×16=120+62+128=310g/mol

310gCa₃(PO₄)₂...120gCa...62gP...128gO₂

100gCa₃(PO₄)₂...xgCa.......ygP.........zgO₂

x=100×120/310=calculeaza

y=100×62/310=calculeaza

z=100×128/310=calculeaza

Na2SO4 sulfat de sodiu

r.at. Na:S:O=2:1:4

r.m Na:S:O=A Cu: A S= 46:32:64=23:16:32

compozitia procentuala

MNa2SO4= 2×23+32+4×16=142g/mol

142Na2SO4.............46gNa.................32gS.................64gO

100gNa2SO4.............xgNa......................ygS....................zgO

x=100×46/142 =calculeaza

y=100×32/142=calculeaza

z=100×640/142=calculeaza

Explicație: