Giải giúp mình B5 ạ
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Bài 5:
Xét ΔABC vuông tại A có
\(\widehat{B}=30^0\)
nên \(AC=\dfrac{1}{2}BC\)
=>BC=12cm
Ta có: ΔABC vuông tại A
mà AM là đường trung tuyến ứng với cạnh huyền BC
nên AM=BC/2=6(cm)
4.
\(A=\left\{1;2;3\right\}\) ; \(B=\left\{0;1;2;3\right\}\) ; \(C=[0;+\infty)\) ; \(D=\left\{\dfrac{1}{2};3\right\}\)
\(\Rightarrow A\subset B\) ; \(A\subset C\); \(B\subset C\) ; \(D\subset C\)
5.
\(A=\left\{\dfrac{1}{2};2\right\}\) ; \(B=\left\{\dfrac{1}{3};2\right\}\)
\(A\cap B=\left\{2\right\}\) ; \(A\cup B=\left\{\dfrac{1}{3};\dfrac{1}{2};2\right\}\)
\(A\backslash B=\left\{\dfrac{1}{2}\right\}\) ; \(B\backslash A=\left\{\dfrac{1}{3}\right\}\)
7.
Các tập con:
\(\varnothing;\left\{a\right\};\left\{b\right\};\left\{c\right\};\left\{d\right\};\left\{a;b\right\};\left\{a;c\right\};\left\{a;d\right\};\left\{b;c\right\};\left\{b;d\right\};\left\{c;d\right\}\)
\(\left\{a;b;c\right\};\left\{a;b;d\right\};\left\{a;c;d\right\};\left\{b;c;d\right\};\left\{a;b;c;d\right\}\)
PTHH : 2Al + 6HCl --> 2AlCl3 + 3H2 ↑ (1)
nAlCl3 = \(\dfrac{m}{M}=\dfrac{13,35}{27+35,5.3}=0.1\left(mol\right)\)
Từ (1) => nHCl = 2nH2 = 0.2 (mol)
=> mHCl = n.M = 0.2 x 36.5 = 7.3 (g)
\(PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{AlCl_3}=\dfrac{m}{M}=\dfrac{13,35}{133,5}=0,1\left(mol\right)\\ Theo.PTHH:n_{HCl}=3.n_{AlCl_3}=3.0,1=0,3\left(mol\right)\\ m_{HCl}=n.M=0,3.36,5=10,95\left(g\right)\)
Bài 3 :
a) nKMnO4 = 31,6/158 = 0,2 (mol)
PTHH : \(2KMnO_4-t^o->K_2MnO_4+MnO_2+O_2\)
Theo pthh : nO2 = 1/2 nKMnO4 = 0,1 (mol)
=> VO2 = 0,1.22,4 = 2,24 (l)
b) PTHH : \(2Mg+O_2-t^o->2MgO\)
Theo pthh : nMgO = 2nO2 = 0,2 (mol)
=> mMgO = 0,2.40 = 8 (g)
Bài 5 :
a) nCu = 0,2 (mol)
nO2 = 0,2 (mol)
PTHH : \(2Cu+O_2-t^o->2CuO\)
Nhận thấy : \(\dfrac{nCu}{2}< nO_2\left(\dfrac{0,2}{2}< 0,2\right)\) => Spu,Cu hết, O2 dư
Theo pthh : nCuO = nCu = 0,2 (mol)
=> mCuO = 0,2.80 = 16 (g)
b) PTHH : \(CuO+H_2-t^o->Cu+H_2O\)
Theo pthh : nH2 = nCuO = 0,2 (mol)
=> VH2 = 0,2.22,4 = 4,48 (l)
Part 1:
1. with
2. does
3. after
4. if
5. where
6. I won't
Part 2:
1B 2F 3A 4E 5C 6D
Part 3:
1. invited
2. will play
3. have just won
4. be cleaned
5. to take
6. beautifully
7. impression
Part 4:
1. read -> reading
2. disappointing -> disappointed
3. environment -> environmental
4. who -> that
Part 5:
1. the bad weather
2. are made to study
3. were good at learning
4. going to the English
a: ĐKXĐ: \(x\notin\left\{10;-10;\sqrt{10};-\sqrt{10}\right\}\)
b: \(A=\dfrac{5x^3+50x+2x^2+20+5x^3-50x-2x^2+20}{\left(x^2-10\right)\left(x^2+10\right)}\cdot\dfrac{x^2-100}{x^2+4}\)
\(=\dfrac{10x^3+40}{\left(x^2-10\right)\left(x^2+10\right)}\cdot\dfrac{x^2-100}{x^2+4}\)
No2 ạ