Đốt cháy 13gam Zn trong bình chứa oxi. Hãy tính:
a. Thể tích O2 dkc phản ứng
b. Tính khối lượng ZnO thu được
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Dạng 3: from... to....= between ... and ...
1. The film starts at 7 pm and ends at 9 pm .
-> The film starts from 7 pm to 9 pm
-> The film start between 7pm and 9 pm
2. We study in the library from 8 am to 10 am.
-> We study between 8 am and 10 am
3. My father works between 9am and 7 pm .
-> My father works from 9 am to 7pm
4. The store opens at 6 am and closes at 8 pm
.-> The store opens from 6 am to 8 am
-> The store opens between 6 am and 8 am
Dang 4: love / like / enjoy + Ving / N = tobe interested in / tobe keen on / tobe fond of + Ving/N
1. I love chatting with friends in my free time.
->I am interested in chatting with friends in my free time
2. He enjoys baking cakes on the weekend.
-> He is keen on baking cakes on the weekend
3. They like making model after a hard-working day.
-> They are interested in making model after a hard-working day
4. He is interested in dancing.
-> He is fond of dancing
Reading
We are happy to announce the 2023 Hien Luong Village Tét Festival will take place on Lê Duẩn Street from January 21 January 29 from 8 a.m to 10 p.m.. Come and celebrate the 2023 Hiền Lương Village Tét Festival next Saturday. This is a free event for everyone. There are lots of activities and types of traditional food to enjoy. Come to Hiền Lương People's Committee and play exciting folk games. Enjoy activities like street music performance and lion dances. Watch the amazing fireworks show on Tét Eve from Hiền Lương Bridge. Enjoy different types of traditional Tết food like bánh chung or candied fruit. Bring your family and friends!
1.The 2023 Hien Luong Festival will last in 9 days. T
2. The festival is only free for children. F
3. There are some activities for children to enjoy. NG
4. Folk games will celebrate at People;s Committee. T
5. You can enjoy music performances from famous Vietnamese singers NG
\(m_{HCl}=219.10\%=21,9\left(g\right)\Rightarrow n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Theo PT: \(n_{ZnCl_2}=n_{H_2}=\dfrac{1}{2}n_{HCl}=0,3\left(mol\right)\)
a, \(V_{H_2}=0,3.24,79=7,437\left(l\right)\)
b, \(m_{ZnCl_2}=0,3.136=40,8\left(g\right)\)
a) (1/3 x²y)(2xy³)
= (1/3 . 2).(x².x).(y.y³)
= 2/3 x³y⁴
Hệ số: 2/3
Phần biến: x³y⁴
Bậc: 7
b) 1/4 x³y .(-2x³y⁴)
= [1/4 . (-2)].(x³.x³).(y.y⁴)
= -1/2 x⁶y⁵
Hệ số: -1/2
Phần biến: x⁶y⁵
Bậc: 11
c) -xy.(2x³y⁴).(-5/4x²y³)
= [-2.(-5/4)].(x.x³.x²).(y.y⁴.y³)
= 5/2 x⁶y⁸
Hệ số: 5/2
Phần biến: x⁶y⁸
Bậc: 14
a chia 5 dư 1 nên \(a=5m+1\left(m\inℕ\right)\)
b chia 5 dư 4 nên \(b=5n+4\left(n\inℕ\right)\)
Do đó \(ab=\left(5m+1\right)\left(5n+4\right)+1\)
\(ab=25mn+20m+5n+4+1\)
\(ab=25mn+20m+5n+5⋮5\)
Ta có đpcm
`#3107.101107`
a.
Số mol của Zn có trong pứ là:
\(\text{n}_{\text{Zn}}=\dfrac{\text{m}_{\text{Zn}}}{\text{M}_{\text{Zn}}}=\dfrac{13}{65}=0,2\left(\text{mol}\right)\)
PTHH: \(2\text{Zn}+\text{O}_2\) \(\underrightarrow{\text{ t}^0\text{ }}\) \(2\text{ZnO}\)
Theo pt: 2 : 1 : 2
`\Rightarrow` n của O2 có trong pứ là `0,1` mol
Thể tích của O2 ở đkc là:
\(\text{V}_{\text{O}_2}=\text{n}_{\text{O}_2}\cdot24,79=0,1\cdot24,79=2,479\left(\text{l}\right)\)
b.
Theo pt: 2 : 1 : 2
`\Rightarrow` n của ZnO thu được sau pứ là `0,2` mol
Khối lượng của ZnO thu được sau pứ là:
\(\text{m}_{\text{ZnO}}=\text{n}_{\text{ZnO}}\cdot\text{M}_{\text{ZnO}}=0,2\cdot\left(16+65\right)=0,2\cdot81=16,2\left(\text{g}\right)\)
Vậy:
a. `2,479` l
b. `16,2` g.
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PT: \(2Zn+O_2\underrightarrow{t^o}2ZnO\)
a, \(n_{O_2}=\dfrac{1}{2}n_{Zn}=0,1\left(mol\right)\Rightarrow V_{O_2}=0,1.24,79=2,479\left(l\right)\)
b, \(n_{ZnO}=n_{Zn}=0,2\left(mol\right)\Rightarrow m_{ZnO}=0,2.81=16,2\left(g\right)\)