3, a, Ta co: S hoa trị II

theo QTHT: + K₂S \(\Rightarrow\) K hóa tri I

+ MgS \(\Rightarrow\) Mg hóa tri II

+ Cr₂S₃ \(\Rightarrow\) Cr hóa tri III

+ CS₂ \(\Rightarrow\) C hóa tri IV

b,Ta co Cl hoa tri I

theo QTHT: + KCl => K hoa tgri I

+ HCl => H hóa tgri I

+ BaCl₂ => Ba hoa tri II

+ AlCl₃ => Al hoa tri III

c, Ta co O hóa tri II

theo QTHT: + Fe₂O₃ => Fe hoa tri III

+ CuO => Cu hoa tri II

+ N₂O₃ => N hoa tri III

+ SO₄ là 1 nhom nguyen tu nên chỉ có hóa tri chung cua nhom ( II)

d, ta co H hoa tri I

theo QTHT: + NH₃ => N hoa tri III

+ C₂H₂ ( hình như bn viet sai đề)

+ HBr => Br hoa tri I

+ H₂S => S hoa tri II

e, ta co CO₃ va SO₄ hoa tri II

theo QTHT: + ZnCO₃ => Zn hoa tri II

+ BaSO₄ => Ba hoa tri II

+ Li₂CO₃ => Li hoa tri I

+ Cr₂(SO₄)₃ => Cr hoa tri III

f, ta co NO₃ va OH hoa tri I

theo QTHT: + NaOH => Na hoa tri I

+ Zn(OH)₂ => Zn hoa tri II

+ AgNO₃ => Ag hoa tri I

+ Al(NO₃)₃ => Al hoa tri III

2, a, CT: Fe_xO_y

Ta co: \(\%Fe=\dfrac{56x}{56x+16y}.100\)

\(\%O=\dfrac{16y}{56x+16y}.100\)

\(\Rightarrow\dfrac{\%Fe}{\%O}=\dfrac{56x}{16y}=\dfrac{72,4}{27,6}\)

\(\Rightarrow\dfrac{x}{y}=\dfrac{362}{483}\approx\dfrac{2}{3}\)

\(\Rightarrow CT:Fe_2O_3\)

b+c, \(n_{Fe}=\dfrac{1}{56}\)

Fe₂O₃ + 3CO \(\rightarrow\) 3CO₂ + 2Fe

de: \(\dfrac{1}{112}\) \(\leftarrow\) \(\leftarrow\dfrac{1}{56}\)

\(m_{Fe_2O_3}=160.\dfrac{1}{112}\approx1,43\left(tan\right)\)

3b, CT: Fe_xO_y = 56x + 16y = 160

theo dề ta co: \(\dfrac{m_{Fe}}{m_O}=\dfrac{7}{3}\)

\(\Rightarrow\dfrac{56x}{16y}=\dfrac{7}{3}\Rightarrow\dfrac{x}{y}=\dfrac{112}{168}=\dfrac{2}{3}\)

\(\Rightarrow CT:Fe_2O_3\)

3a, bn lm tuong tự câu b

A=3+3²+3³+...+3²²

A=(3+3²)+(3³+3⁴)+...+(3²¹+3²²)

A=3(1+3)+3³(1+3)+...+3²¹(1+3)

A=3.4+3³.4+...+3²¹.4

A=4(3+3³+...+3²¹) chia hết cho 4 - đpcm

Nhấn đúng nha

3a

2b

4, a, CuCO₃ \(\underrightarrow{t^o}\) CuO + CO₂

Cu(OH)₂ \(\underrightarrow{t^o}\) CuO + H₂O

theo ĐLBTKL:

\(m_{CuCO_3}+m_{Cu\left(OH\right)_2}=m_{CO_2}+m_{H_2O}+m_{CuO}\)

\(\Rightarrow m_{tc}=120-\left(80+22+9\right)=9g\)

\(\Rightarrow\%m_{tc}=\dfrac{9}{120}.100\%=7,5\%\)

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{14,7}{98}=0,15\left(mol\right)\)

Ta co: \(\dfrac{0,2}{1}>\dfrac{0,15}{1}\Rightarrow Mg\) dư

Mg + H₂SO₄ \(\rightarrow\) MgSO₄ + H₂

de: 0,2 0,15

pu: 0,15 0,15 0,15 0,15

spu: 0,05 0 0,15 0,15

a, \(V_{H_2}=0,15.22,4=3,36l\)

b, \(m_{Mg\left(dư\right)}=0,05.24=1,2g\)

a, SO₂

1, CT: AO

\(m_{A\left(OH\right)_2}=\dfrac{200.8,55}{100}=17,1g\)

AO + H₂O \(\rightarrow\) A(OH)₂

pt: A + 16 A + 32 + 2

de: 15,3 17,1

Ta co:\(17,1\left(A+16\right)=15,3\left(A+32+2\right)\)

\(\Leftrightarrow17,1A+273,6=15,3A+520,2\)

\(\Leftrightarrow1,8A=246,6\Rightarrow A=137\)

\(\Rightarrow CT:BaO\)

TT: m₁ = 500g = 0,5kg ; t^o₁ = 0^oC

m₂ = 2kg ; t^o₂ = 40^oC ; t^o = 5^oC

c= 4200 J/kg.K => n = ?

giai:

nhiet luong coc nc toa:

Q_t = m₂ . c .(t^o₂ - t^o) = 294000J

nhiet luong cuc da thu vao:

Q_th = m . c .( t^o - t^o₁) = 21000m

theo PTCBN: Q_t = Q_th

\(\Leftrightarrow21000m=294000\Rightarrow m=14kg\)

so cuc da can phai bo vao coc nc la:

\(n=\dfrac{m}{m_1}=28\) (cục)

vì mỗi lần bỏ 1 cuc nên phải bo 28 lần