Giúp vs ạ đag cần gấp
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2KClO3 -> (t°, MnO2) 2KCl + 3O2
3O2 -> (UV) 2O3
O3 + 2Ag -> Ag2O + O2
4Na + O2 -> (t°) 2Na2O
Na2O + H2O -> 2NaOH
2NaOH + Cl2 -> NaCl + NaClO + H2O
2NaCl -> (đpnc) 2Na + Cl2
H2S + 4Cl2 + 4H2O -> H2SO4 + 8HCl
\(2KClO_3\rightarrow\left(t^o,MnO_2\right)2KCl+3O_2\)
\(3O_2\rightarrow\left(tia.UV\right)2O_3\)
\(2Ag+O_3\rightarrow Ag_2O+O_2\)
\(4Na+O_2\rightarrow2Na_2O\)
\(Na_2O+H_2O\rightarrow2NaOH\)
\(NaOH+HCl\rightarrow NaCl+H_2O\)
\(2NaCl\rightarrow\left(đp\right)2Na+Cl_2\)
\(Cl_2+2H_2O+SO_2\rightarrow H_2SO_4+2HCl\)
1 D
2 C
3 C
4 C
5 A
6 B
7 A
8 C
9 B
10 D
11 C
12 A
13 B
14 C
15 B
16 D
17 B
18 D
19 A
20 A
a) \(p=d\cdot h=150\cdot10^{-4}\cdot10000=150Pa\)
b) \(F_A=d\cdot V=10000\cdot400\cdot10^{-6}=4N\)
*Theo mình là
D.Cung cấp chất dinh dưỡng cho cây trồng
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a) Ta có: \(P=\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{3\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}+2}{\sqrt{a}+1}\right)\)
\(=\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}+\dfrac{a-1}{\sqrt{a}}\cdot\dfrac{3\sqrt{a}\left(\sqrt{a}+1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}+\dfrac{3a+3\sqrt{a}-\left(a-\sqrt{a}+2\sqrt{a}-2\right)}{\sqrt{a}}\)
\(=2+\dfrac{3a+3\sqrt{a}-a+\sqrt{a}-2\sqrt{a}+2}{\sqrt{a}}\)
\(=\dfrac{2\sqrt{a}+2a+2\sqrt{a}+2}{\sqrt{a}}\)
\(=\dfrac{2\left(a+2\sqrt{a}+1\right)}{\sqrt{a}}\)
\(=\dfrac{2\left(\sqrt{a}+1\right)^2}{\sqrt{a}}\)
b) Ta có: \(P-6=\dfrac{2\left(\sqrt{a}+1\right)^2-6\sqrt{a}}{\sqrt{a}}\)
\(=\dfrac{2a+4\sqrt{a}+2-6\sqrt{a}}{\sqrt{a}}\)
\(=\dfrac{2\left(a-\sqrt{a}+1\right)}{\sqrt{a}}>0\forall a\) thỏa mãn ĐKXĐ
hay P>6