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123+123*2/3=123+82=225

\(123+123.\frac{2}{3}\)

\(=123+82\)

\(=205\)

205 nhé

\(\dfrac{x-1}{3}=\dfrac{2-x}{-2}\)

⇔ \(\dfrac{x-1}{3}=\dfrac{x-2}{2}\)

⇔ \(3x-6-2x+2=0\)

⇔ \(x-4=0\)

⇒ \(x=4\)

=>-2x+2=6-3x

=>-2x+3x=6-2

=>x=4

1 he didn't have to study for his exam

2 she weren't lazy, she could pass the exam

3 If my brother had left the car keys, I could have picked him up at the station

4 he pays me tonight, I will have enough money to buy a car

5 he didn't smoke too much, he could get rid of his cough

6 I hadn't lost my key, I wouldn't have had to pound on the door.....

7 she weren't shy, she would enjoy the party

8 I get a work permit, I will stay for another month

9 he took some exercises, he wouldn't be so unhealthy

10 those people had been prepared  to face the floods, the consequence wouldn't have been disastrous

11 she starts working hard now, she won't be able to pass the final test

12 you are patient, you won't get success

13 somebody waters these flowers. they will die

14 it doesn't stop raining, we won't go out

15 hhe hadn't drunk alcohol, he would have passed the contest

\(=11\cdot\dfrac{16}{55}\cdot\dfrac{15}{8}=\dfrac{11\cdot5}{55}\cdot\dfrac{16}{8}\cdot3=2\cdot3=6\)

\(MgS,Mg\left(NO_3\right)_2,MgCO_3,Mg_3\left(PO_4\right)_2\)

\(Fe_2S_3,Fe\left(NO_3\right)_3,Fe_2\left(CO_3\right)_3,FePO_4\)

\(\left(NH_4\right)_2S,NH_4NO_3,\left(NH_4\right)_2CO_3,\left(NH_4\right)_3PO_4\)

mgs, mg(no3)2, mgco3, mg3(po4)2, 

1. Gọi số mol Fe trong hỗn hợp là a

m = mNa + mFe 

+) Hỗn hợp tác dụng hết với HCl: 

2Na + 2HCl → 2NaCl + H2↑

Fe + 2HCl → FeCl2 + H2↑

 a----------------->a

Dung dịch thu được gồm: NaCl, FeCl2, HCl (có thể còn dư)

+) Dung dịch thu được tác dụng với Ba(OH)2 dư:

2HCl + Ba(OH)2 → BaCl2 + 2H2O

FeCl2 + Ba(OH)2 → BaCl2 + Fe(OH)2↓

a--------------------------------------->a

Kết tủa: Fe(OH)2

+) Nung kết tủa đến khối lượng không đổi:

4Fe(OH)2 + O2 → 2Fe2O3 + 4H2O

a---------------------------a/2

mcr = m = mFe2O3 = a/2 . 160 = 80a

\(m_{Fe}=\dfrac{56a}{80a}.100=70\%\)

%mNa=100%−70%=30%

2/

CaCO3 ---^_to--> CaO + CO2
x----------------------x
MgCO3 ---^_to--> MgO + CO2
y----------------------y
m_{ban đầu}= 2.m_{sau nung}
=> 100x+84y = 2 .(56x+40y)
=> 12x = 4y
=> \(\dfrac{n_{CaCO_3}}{n_{MgCO_3}}=\dfrac{1}{3}\)

Đặt n _CaCO3 = 1mol =>  n _MgCO3 = 3mol 
%CaCO3=\(\dfrac{100}{100+84.3}.100\)=28,41%
%MgCO3=100-28,41=71,59%

XIN CẢM ƠN MN TRƯỚC NHƯNG MIK CẦN GẤP NHA ^^0

a: \(\dfrac{1.2}{3.24}=\dfrac{120}{324}\)

b: \(\dfrac{2+\dfrac{1}{5}}{\dfrac{3}{4}}=\dfrac{11}{5}:\dfrac{3}{4}=\dfrac{44}{15}\)