nCO2=v/22,4=6,72/22,4=0,3(mol)
a)PTHH: CO2 + Ca(OH)2 ➝ CaCO3 + H2O(1)
             0,3    ➝   0,3         ➝    0,3               (mol)

b) Từ 1 ⇒nCa(OH)2=nCaCO3=nCO2=0,3(mol)
⇒mCa(OH)2=n.M=0,3.74=22,2(g)
⇒mCaCO3=n.M=0,3.100=30(g)

nNa2O=m/M=15,5/62=0,25(mol)
PT: Na2O + H2O ➝ 2NaOH (1)
        0,25              ➝    0,5    (mol)
a)Từ (1)=> C_M(NaOH)= n/Vdd=0,5/0,5=1(M)
b) PT: 2NaOH + H2SO4  ➝ Na2SO4 + 2H2O (2)
             0,5     ➝   0,25                                    (mol)
Từ(2)=>mH2SO4=n.M=0,25.98=24,5(g)
=>mddH2SO4=(mctH2SO4.100%)/C%=(24,5.100%)/20%
=122,5(g)
=>VddH2SO4=mdd/D=122,5/1,14=108(ml)=0,108(l)

\(\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right).\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{\sqrt{xy}}{\sqrt{x}-\sqrt{y}}\)

PTHH: H2SO4 + 2NaOH -> Na2SO4 + 2H2O(1)
              0,12    ->  0,24    ->   0,12                 (mol)
nH2SO4=Vdd.C(M)=0,06.2=0,12(mol)
-VddNaOH=n/C(M)=0,24/1=0,24(l)
VddSPỨ =VddH2SO4 + VddNaOH =0,06+0,24=0,3(l)
-C(M)ddA = C(M)Na2SO4=n/Vdd=0,12/0,3=0,4(M)

Đà Nẵng có bão, mai tui được nghỉ mn ơi :DD

\(\left(3-a\right)^2-\sqrt{0,2}.\sqrt{180a^2}\)

nHCl=Vdd.C_M=0,61=0,6(mol)
PTHH: Al2O3 + 6HCl ➝ 2AlCl₃ + 3H2O (1)
                      0,6  ➝ 0,2                    (mol)
Từ(1) => nAlCl3=0,2(mol)
          => mAlCl3=n.M=0,2.133,5=26,7(g)

c) B.C = \(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}\) \(=\) \(1-\dfrac{2}{\sqrt{x}+1}\)

Để B.C ∈ Z thì \(\dfrac{2}{\sqrt{x}+1}\)∈ Z.
\(\Leftrightarrow2⋮\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow\sqrt{x}+1\in\left\{1;-1;2;-2\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{0;-2;1;-3\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{0;1\right\}\)
\(\Leftrightarrow x\in\left\{0;1\right\}\)
Để B.C ∈ Z thì x ∈ \(\left\{0;1\right\}\)
Mà x > 0; x ≠ 1 nên B.C không thể nhận giá trị nguyên.