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Kimyo fanidan 9-sinflar uchun yozma ish savollari
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| bet | 6/6 | | Sana | 27.11.2023 | | Hajmi | 103 Kb. | | #106687 |
Bog'liq 1- kimyo fec985cc-3580-49d3-8276-efd38823e807, 1-Mavzu. Kiberxavfsizlikga oid milliy va xorijiy me’yoriy-huquqi-kompy.info, ddddddd, 1-3 labaratoriya ishi, BILOLIDDIN (Tizimlardan), 12-ma\'ruza KOMPYUTER LOYIHALSH, Laboratoriya 3 ShirinKimyo fanidan 9-sinflar uchun yozma ish savollari
III-variant
1-masala 50 ml 0.5 M li kalsiy xlorid eritmasini 0.2 M li natriy karbonat eritmasi bilan to`liq reaksiyaga kirishishi uchun qancha hajm natriy korbanat kerak?
2-masala Massasi 20 g bo`lgan temir plastinka СuSO4 eritmasiga tushurilda. Plastinka massasi 22 g ga teng bo`lgan vaqtda eritmaga temirning qanday massasi o`tgan?
3-masala Tabiiy misning 73% Cu va 27% Cu izotoplarining aralashmasidan iborat bo`lsa uning o`rtacha molekulyar massasini aniqlang?
4-masala Quyidgi oksidlanish qaytarilish reaksiyasini davom ettiring. Koeffisientlar yig`indisini toping?
FeSO4 +KMnO4+H2SO4 =
Savollar yangi tahrirdagi DTS asosda tuzilgan
Monitoring guruhi N.Orziyeva
N.Usmonova
A.Ravshanova
D.Murodova
S.Halimova
S.Halimova
9-sinflar uchun kimyo fanidan olimpiada test javoblari
I-variant
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18.D= D=
19.Modda konsentratsiyasining vaqtga nisbati
20. To`g`ri reaksiya tezligining teskari reaksiya tezligiga teng bo`lishi kimyoviy muvozanat deyiladi, uni siljitish yo`llari temperatura,bosim, konsentratsiya
21.Na2O*CaO*6SiO2
22.qarama- qarshi tomonga boradigan reaksiya
23. FeS2
24. 6 ta
25. 1 mol (Fe3O4)________ 168 g (Fe)
0.5 mol __________ X= 84 g
9-sinflar uchun kimyo fanidan olimpiada test javoblari
II-variant
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18. 2 Xil . Karbonatlar,Gidrokarbonatlar
19. Katodda natriy, Anodda xlor
20. 27 g(Al)________ 6.02*1023
13.5 __________ X=3.01*1023
21.56 g(Fe)________ 64 g(Cu)
2.8 _________ X=3.2
22. Aluminiy
23.FeS2
24. 1 mol (Fe3O4)________ 168 g (Fe)
0.5 mol __________ X= 84 g
25 Modda konsentratsiyasining vaqtga nisbati
9-sinflar uchun kimyo fanidan olimpiada test javoblari
III-variant
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18. FeS2
19. 6 ta
20. 6ta
21. 1 mol (Fe3O4)________ 168 g (Fe)
0.5 mol __________ X= 84 g
22.Ca(OH)2
23.Ca3(PO4)3
24. Na2O*CaO*6SiO2
25.4Fe+6H2O+3O2=4Fe(OH)3
Kimyo fanidan 9-sinflar uchun yozma ish savollarining javoblari
I-variant
1. 18(H2O) _____ 2 22.4 ______ 28
7.2 ______ X=0.8(H2) 4.48 ______ X=5.6(N)
5.6+0.8=6.4 12.8-6.4=6.4(O)
N/H/O=
Javob NH4NO2
2.Ca3(PO4)2+3H2SO4=3CaSO4+2H3PO4
196(H3PO4) ______ 310(Ca3(PO4)2) 310 _______ 100%
195 _______ X=308.4 308,4 ______ X=99.5%
Javob 99.5
3.M(Cu)=
Javob 63.54
4.CaCO3____CaO+CO2 CO2+2NaOH=Na2CO3+H2O
NaHCO3 +HCl=NaCL+H2O+CO2 Na2CO3+H2O+ CO2=2NaHCO3
Kimyo fanidan 9-sinflar uchun yozma ish savollarining javoblari
II-variant
1.CaCO3______ CaO+CO2
100(CaCO3 )________ 56(CaO) 60__________ 100%
57.1=X _________ 32 57.1 _________ X=95.2%
Javob 95.2 %
2.CuSO4+Fe=FeSO4+Cu
56 64
64-56=8 22-20=2
8 _______ 56(Fe)
2 _______ X=14 g Javob 14 g (Fe)
3.Na2CO3*nH2O
100%-62.94%=37.06%
37.06% (Na2CO3)_________ 62.94%(H2O)
106 ___________ X=180 g (H2O)
n=
Javob Na2CO3*10H2O
4.2Fe+O2=2FeO
FeO+H2SO4=FeSO4+H2O
10FeSO4+2KMnO4+8H2SO4=5Fe2(SO4)3+K2SO4+2MnSO4+8H2O
Fe2(SO4)3+6NaOH=3Na2SO4+2Fe(OH)3
2Fe(OH)3____ Fe2O3+3H2O
Fe2O3+3H2=2Fe+3H2O
Kimyo fanidan 9-sinflar uchun yozma ish savollarining javoblari
III-variant
1.V2=
2. CuSO4+Fe=FeSO4+Cu
56 64
64-56=8 22-20=2
8 _______ 56(Fe)
2 _______ X=14 g Javob 14 g (Fe)
3.M(Cu)=
Javob 63.54
4. 10FeSO4+2KMnO4+8H2SO4=5Fe2(SO4)3+K2SO4+2MnSO4+8H2O
Javob: 36
@ziyoilm-канали орқали сизга ёрдам берамиз! Мурожаат учун: @quvonchbek_karimov
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