giúp mik với ạ II nha
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DA
2
MH
12 tháng 1 2022
\(\dfrac{x-1}{3}=\dfrac{2-x}{-2}\)
⇔ \(\dfrac{x-1}{3}=\dfrac{x-2}{2}\)
⇔ \(3x-6-2x+2=0\)
⇔ \(x-4=0\)
⇒ \(x=4\)
CT
1
DT
Đỗ Thanh Hải
CTVVIP
22 tháng 10 2021
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
22 tháng 2 2022
\(=11\cdot\dfrac{16}{55}\cdot\dfrac{15}{8}=\dfrac{11\cdot5}{55}\cdot\dfrac{16}{8}\cdot3=2\cdot3=6\)
CH
16 tháng 8 2021
\(MgS,Mg\left(NO_3\right)_2,MgCO_3,Mg_3\left(PO_4\right)_2\)
\(Fe_2S_3,Fe\left(NO_3\right)_3,Fe_2\left(CO_3\right)_3,FePO_4\)
\(\left(NH_4\right)_2S,NH_4NO_3,\left(NH_4\right)_2CO_3,\left(NH_4\right)_3PO_4\)
TP
25 tháng 7 2021
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%
TP
25 tháng 7 2021
2/
CaCO3 ---to--> CaO + CO2
x----------------------x
MgCO3 ---to--> MgO + CO2
y----------------------y
mban đầu= 2.msau 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%
24 tháng 10 2021
a: \(\dfrac{1.2}{3.24}=\dfrac{120}{324}\)
b: \(\dfrac{2+\dfrac{1}{5}}{\dfrac{3}{4}}=\dfrac{11}{5}:\dfrac{3}{4}=\dfrac{44}{15}\)
II
electricity
pollution
environmental
renewable
effective
convenient
Alternative
comsuption
easily
dangerous
bạn giải III luôn đc ko