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15 tháng 10 2021
a, \(n_{KClO_3}=\dfrac{36,75}{122,5}=0,3\left(mol\right)\)
PTHH: 2KClO3 ---to→ 2KCl + 3O2
Mol: 0,3 0,3 0,45
\(m_{KCl}=0,3.74,5=22,35\left(g\right)\)
\(V_{O_2}=0,45.22,4=10,08\left(l\right)\)
b,
PTHH: 4Fe + 3O2 ---to→ 2Fe2O3
Mol: 0,6 0,45
\(m_{Fe}=0,6.56=33,6\left(g\right)\)
12 tháng 3 2023
R=1/2CD=a
h=AD=2a
S1=Sxq=2*pi*r*h=2*pi*a*2a=4*pi*a^2
S2=Stp=2*pi*r^2+2*pi*r*h
=2*pi*a^2+2*pi*a*2a
=6*pi*a^2
>S1/S2=2/3
AT
29 tháng 7 2021
\(P=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\left(x\ge0,x\ne4\right)\)
\(=\dfrac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)
\(=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
c) \(P=\dfrac{4}{3}\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}=\dfrac{4}{3}\Rightarrow3\sqrt{x}=4\sqrt{x}-8\Rightarrow\sqrt{x}=8\Rightarrow x=64\)
mk cần gấp, giúp mk vói ạ
a,A=\(\left(2+\dfrac{2+\sqrt{3}}{\sqrt{3}+1}\right)\left(2-\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)\)
=\(\left(\dfrac{2\left(\sqrt{3}+1\right)+2+\sqrt{3}}{\sqrt{3}+1}\right)\left(\dfrac{2\left(\sqrt{3}-1\right)-3+\sqrt{3}}{\sqrt{3}-1}\right)\)
=\(\left(\dfrac{2\sqrt{3}+2+2+\sqrt{3}}{\sqrt{3}+1}\right)\left(\dfrac{2\sqrt{3}-2-3+\sqrt{3}}{\sqrt{3}-1}\right)\)
=\(\dfrac{3\sqrt{3}+4}{\sqrt{3}+1}\times\dfrac{3\sqrt{3}-5}{\sqrt{3}-1}\)
=\(\dfrac{\left(3\sqrt{3}+4\right)\left(3\sqrt{3}-5\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
=\(\dfrac{27-15\sqrt{3}+12\sqrt{3}-20}{3-1}\)
=\(\dfrac{7-3\sqrt{3}}{2}\)
b,B=\(\left(\dfrac{\sqrt{a}}{a-\sqrt{ab}}-\dfrac{\sqrt{a}}{\sqrt{ab}-b}\right)\left(a\sqrt{a}-b\sqrt{a}\right)\)
=\(\left(\dfrac{\sqrt{a}.\sqrt{b}-\sqrt{a}.\sqrt{a}}{\sqrt{ab}.\left(\sqrt{a}-\sqrt{b}\right)}\right).\sqrt{a}.\left(a-b\right)\)
=\(\left(\dfrac{\sqrt{ab}-a}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\right).\sqrt{a}\left(a-b\right)\)
=\(\left(\dfrac{-\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}\right).\sqrt{a}\left(a-b\right)\)
=\(\dfrac{-1}{\sqrt{b}}.\sqrt{a}\left(a-b\right)\)
=\(\dfrac{-\sqrt{a}\left(a-b\right)}{\sqrt{b}}\)