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The Eiffel Tower is located in Paris, France. It was constructed between 1887 and 1889 as anentrance way to the 1889 World’s Fair and to celebrate the 100thanniversary of the French Revolution. The Tower was opened to visitors on May 6th, 1889.
Gustave Eiffel’s design was chosen from among 107 designs that were submitted to the World’s Fair design competition. However, many Parisians, especially artists, did not like his design and protested the tower’s construction. They thought it would be an eyesore, but once it was built, most Parisians soon loved the tower.
The tower is made of iron and weighs over 10,000 tons. It is 324 meters tall, including an antenna at its top, and has a staircase with 1,665 steps. There are also elevators to take visitors to the top platform where there is a panoramic view of Paris. The original elevators, now computerized, are still in use. Over 60 tons of paint are applied to the tower every seven years to keep it from rusting.
The Eiffel Tower has become a symbol of Paris. It is the most recognized monument in Europe mainly because many people think it is an architectural masterpiece. Over 250 million people have visited it since May of 1889.
70.66. How long did it take to build The Eiffel Tower?(1 Point)A. around 2 years B. around 6 years C. 100 yearsD. 60 years
71.67. Who designed The Eiffel Tower? (1 Point)A. the Parisians B. the French RevolutionC. Gustave EiffelD. the World’s Fair
72.68. Why do they repaint The Eiffel Tower frequently?(1 Point)A. to keep it from rustingB. to make it more beautifulC. to attract visitorsD. to display it in the World’s Fair design competition
73.69. How many people have visited The Eiffel Tower until now?(1 Point)A. about 10.000 peopleB. about 324 million peopleC. more than 250 million peopleD. about 189 million people
74.70. What is the major reason for The Eiffel Tower to be a well-known tourist attraction in Europe?(1 Point)A. Because of its history.B. Because it is computerized.C. Because it has architectural beauty.D. Because it is repainted frequently.
n KClO3 = 4,9/122,5 = 0,04(mol)
$2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2$
n O2 = 3/2 n KClO3 = 0,06(mol)
X cho vào HCl thấy thoát ra khí chứng tỏ X chứa R dư
Gọi n là hóa trị của R
n H2 = 1,344/22,4 = 0,06(mol)
$4R + nO_2 \xrightarrow{t^o} 2R_2O_n$
$2R + 2nHCl \to 2RCl_n + nH_2$
n R = 4/n n O2 + 2/n n H2 = 0,36/n(mol)
Bảo toàn khối lượng :
=> m R = m X - m O2 = 6,24 - 0,06.32 = 4,32(gam)
Suy ra :
0,36/n . R = 4,32
=> R = 12n
Với n = 2 thì R = 24(Magie)
\(MgCl_2+2KOH\rightarrow Mg\left(OH\right)_2+2KCl\)
\(Cu\left(OH\right)_2+2HCl\rightarrow CuCl_2+2H_2O\)
1) MgCl2 + 2KOH → Mg(OH)2 + 2KCl
2) Cu(OH)2 + 2HCl → CuCl2 + 2H2O
PTHH: Fe + CuSO4 → FeSO4 + Cu
Mol: x x x x
Theo ĐLBTKL,ta có: 56x+160x = 152x + 64x
⇔ 160x - 152x = 64x - 56x = m+16-m=16
⇔ 8x = 16
⇔ x=2
⇒ m=mFe = 56.2 = 112 (g)
Vậy m=112 g
\(Fe+CuSO_4 \to FeSO_4+Cu\\ n_{Fe}=a(mol)\\ n_{Cu}=a(mol)\\ m+16=64a\\ \to 1,6=64a-56a\\ a=0,2(mol)\\ m_{Fe}=0,2.56=11,2(g)\)
a)
n CaO = n O = 20.20%/16 = 0,25(mol)
%m CaO = 0,25.56/20 .100% = 70%
%m Ca = 100%- 70% = 30%
b)
$Ca + 2H_2O \to Ca(OH)_2 + H_2$
n H2 =n Ca = 20.30%/40 = 0,15(mol)
V = 0,15.22 4 = 3,36(lít)
n Fe3O4 = 23,2/232 = 0,1(mol)
=> n Fe(trong chất rắn) = 0,1.3 = 0,3(mol)
=> m = 0,3.56/78,9474% = 21,28(gam)
c)
$Fe_3O_4 + 4H_2 \xrightarrow{t^o} 3Fe + 4H_2O$
n Fe3O4 pư = 1/4 n H2 = 0,0375(mol)
H = 0,0375/0,1 .100% = 37,5%
\(n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\)
\(n_{HCl}=0,4.1=0,4\left(mol\right)\)
\(CuO+2HCl\rightarrow CuCl_2+H_2O\)
0,15 ---> 0,3 -----> 0,15 ----> 0,15
Xét: \(\dfrac{0,15}{1}< \dfrac{0,4}{2}\) => axit dư
a. HCl dư sau phản ứng và \(n_{HCl.dư}=0,4-0,3=0,1\left(mol\right)\)
b.
Trong dung dịch A có:
\(m_{HCl}=0,1.36,5=3,65\left(g\right)\)
\(m_{CuCl_2}=0,15.135=20,25\left(g\right)\\ m_{H_2O}=0,15.18=2,7\left(g\right)\)
\(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\\ n_{O_2}=\dfrac{22,4}{22,4}=1\left(mol\right)\\ 4P+5O_2\underrightarrow{^{to}}2P_2O_5\\ Vì:\dfrac{0,1}{4}< \dfrac{1}{5}\Rightarrow O_2dư\\ n_{O_2\left(dư\right)}=1-\dfrac{5}{4}.0,1=0,875\left(mol\right)\\ m_{O_2}=0,875.32=28\left(g\right)\\ n_{P_2O_5}=\dfrac{2}{4}.0,1=0,05\left(mol\right)\\ m_{P_2O_5}=0,05.142=7,1\left(g\right)\)
\(n_P=\dfrac{3,1}{31}=0,1\left(mol\right);n_{O_2}=\dfrac{22,4}{22,4}=1\left(mol\right)\)
PTHH: 4P + 5O2 ---to→ 2P2O5
Mol: 0,1 0,125 0,05
Ta có: \(\dfrac{0,1}{4}< \dfrac{1}{5}\) ⇒ P hết, O2 dư
\(m_{O_2dư}=\left(1-0,125\right).32=28\left(g\right)\)
\(m_{P_2O_5}=0,05.142=7,1\left(g\right)\)
ví dụ fe tác dụng với hcl dư tạo thành fecl2 và h2
=> nfecl2 và nh2 sẽ được tính theo fe vì lượng fe pư hết, hcl dư đó bạn
và lượng hcl pư hết cũng đc tính theo lượng fe: nhcl pư= 2nfe
nhcl dư=nhcl ban đầu- nhcl pư hếT
TÓM LẠI TRONG BÀI TOÁN ĐỀ CHO 1 CHẤT DƯ THÌ CHẤT CÒN LẠI SẼ PƯ HẾT, VÀ CÁC CHẤT THU ĐƯỢC TÍNH THEO CHẤT PƯ HẾT ĐÓ BẠN!
mong bạn sẽ hiêu
tứ là những chất tham gia sau khi phản ứng vẫn chưa phản ứng hết thì gọi là dư