1. Мa’lum miqdordagi rux sulfatga 40 LI Naoh eritmasi qo’shilganda 19,8 g cho’kma tushdi va 0,2 mol Na

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1.Мa’lum miqdordagi rux sulfatga 40 % li NaOH eritmasi qo’shilganda 19,8 g cho’kma tushdi va 0,2 mol Na₂[Zn(OH)₄] hosil bo’ldi. Sarflangan ishqor eritmasining massasini (g) hisoblang
.A) 120 B) 140 C) 100 D) 80
ZnSO₄+2NaOH=Na₂SO₄+Zn(OH)₄ 80 g--------------99 g 16 g=x-------------19.8gZnSO₄+4NaOH= Na₂SO₄+Na₂[Zn(OH)₄] 160g-----------------1 mol 32 g=x-----------------0.2 m32+16=48 g 48 g----------40%
120 g=x-------100% javob:120
2. 1.5Nli (p=0,98g/ml) H ₃ PO ₄ eritmasing foiz konsentratsiya sini aniqlang.
Masalani yechilishi. Ushbu masalani normal konsentratsiyadan foiz konsentratsiyaga otish formulasidanvfoydalanib oson ishlashimiz mumkun:
M(H_3.PO ₄ ) 98
E( H₃ PO ₄) =---------------= ------=32.67
n (H) 3

C_{N ×}E. 1.5×32.67
C%= ---------= -------------=5%
p × 10. 0.98×10
Javob.5%
3. 10%li (p=1.025g/ml) H ₂ SO ₃ eritmasining normal konsentratsiyasini aniqlash.
Masalaning yechilishi: Avval H _2. SO ₃. ning ekvivalent massasining topib olamiz.
M_k.ta
E _k.ta=---------
n(H)
M( H_2. SO₃ ) 82
E(H ₂ SO ₃ )=-----------------=----------=41
n (H) 2
Ushbumasalanifoizkonsentratsiyadannormal konsentratsiyaga otishformulasidan foydalanib oson ishimimiz mumkun.
C%×p×10. 10×1.025×10
C _N =-------------=--------------------------2.5N
E. 41
Javob2.5
4.76 g xrom(III) –oksid namunasi vodorod yordamida qisman qaytarildi. Olingan qattiq qoldiq massasi 64 g bo‘lsa, sarflangan vodorod hajmini (1,n.sh.) hisoblang.
A) 33,6 B) 44,8 C) 16,8 D) 89,6
Cr₂O₃ + 3H₂= 2Cr+ 3H₂O152x gr---67.2x l---104x grCr₂O₃------Cr₂O₃152y-------152y152x+152y=76104x+152y=64X=0.25 y=0.2567.2x=67.2•0.25=16.8 lJavob:16.8 l3.
5 СаСО₃ va МgСО₃ to ‘liq parchalanganda olingan qattiq qoldiq massasi dastlabki aralashma massasining yarmini tashkil etdi. 2 mol shunday aralashmadagi СаСО₃ massasini (g) hisoblang.A) 125 B) 75 C) 50 D) 100CaCO₃=CaO+CO₂100x-----56xMgCO₃=MgO+CO₂84y------40yX+y=2 (100x+84y)/2=56x+40yX=0.5 y=1.5 100x=100*0.5=50 gJavob:50