\(9x^2y^2+15x^2y-21xy^2\\ =3xy\left(3xy+5x-7y\right)\)

hình như phải là "thưa thốt" chứ không phải là "thưa thớt" nha bạn

1.B 2.A 3.C 4.A,D 5.D 6.A 7.C 8.C 9.D 10.B

a. Cu, Fe

b.Cu,CuO

c.CaCl2

\(n_{Fe}=0,2\left(mol\right)\)

\(n_{HCl}=1.0,3=0,3\left(mol\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

0,2-------------------0,2-----0,2

\(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

\(m_{FeCl_2}=0,2.127=25,4\left(g\right)\)

\(m_{HCl_{dư}}=\left(0,3-0,2\right).36,5=3,65\left(g\right)\)

\(M_O=16\Rightarrow16.2,5=40\Rightarrow Ca\)

a.\(\dfrac{1}{4}+\dfrac{15}{16}:\dfrac{3}{8}x\dfrac{2}{3}=\dfrac{1}{4}+\dfrac{5}{2}x\dfrac{2}{3}=\dfrac{1}{4}+\dfrac{5}{3}=\dfrac{23}{12}\)

b.

+ Với a=58,47, ta có:

A=4,25.(58,47+41,53)-125=300

+Với A=53,3, ta có:

53,5 = 4,25.(a+41,53)-125 =>a=0,47

AKI=AHI

AKC=AHB

KBI=HCI

KBC=HCB

CaO

a.\(Fe_2O_3+6HCl\rightarrow2FeCl_2+3H_2O\)

b.\(Na_2O+H_2O\rightarrow2NaOH\)

c.\(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_3+H_2O\)

d.\(Al_2O_3+6HNO_3\rightarrow2Al\left(NO_3\right)_3+3H_2O\)

e.\(3FeO+2H_3PO_4\rightarrow Fe_3\left(PO_4\right)_2+3H_2O\)

g.\(3Cu\left(NO_3\right)_2+2Al\rightarrow2Al\left(NO_3\right)_3+3Cu\)