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HT
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)
\(PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{AlCl_3}=\dfrac{m}{M}=\dfrac{13,35}{133,5}=0,1\left(mol\right)\\ Theo.PTHH:n_{HCl}=3.n_{AlCl_3}=3.0,1=0,3\left(mol\right)\\ m_{HCl}=n.M=0,3.36,5=10,95\left(g\right)\)
a) \(\dfrac{3}{4}+\dfrac{9}{5}\div\dfrac{3}{2}-1=\dfrac{3}{4}+\dfrac{18}{15}-1=\dfrac{39}{20}-1=\dfrac{19}{20}\)
b) \(\dfrac{6}{7}\cdot\dfrac{8}{13}+\dfrac{6}{13}\cdot\dfrac{9}{7}-\dfrac{4}{13}\cdot\dfrac{6}{7}=\dfrac{48}{91}+\dfrac{54}{91}-\dfrac{24}{91}=\dfrac{48+51-24}{91}=\dfrac{78}{91}=\dfrac{6}{7}\)
c) \(\dfrac{-3}{7}+\left(\dfrac{3}{-7}-\dfrac{3}{-5}\right)\)\(=\dfrac{-3}{7}+\left(\dfrac{-3}{7}-\dfrac{-3}{5}\right)=\dfrac{-3}{7}+\dfrac{6}{35}=-\dfrac{9}{35}\)
a) \(=3\left(x-3y\right)\)
b) \(=5xy\left(3x-2y\right)\)
c) \(=5y\left(x+2y\right)+2\left(x+2y\right)=\left(x+2y\right)\left(5y+2\right)\)
d) \(=\left(9x^2+6x+1\right)-4y^2=\left(3x+1\right)^2-4y^2=\left(3x+1-2y\right)\left(3x+1+2y\right)\)
e) \(=x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(x+3\right)\)
g) \(=x\left(x-2\right)-\left(x-2\right)=\left(x-2\right)\left(x-1\right)\)
a) \(3x-9y\)
\(=3\left(x-3y\right)\)
b) \(15x^2y-10xy^2\)
\(=5xy\left(3x-2y\right)\)
c) \(5xy+10y^2+2x+4y\)
\(=\left(x+2y\right)\left(5y+2\right)\)
d) \(9x^2-4x^2+6x+1\)
\(=\left(3x+1-2y\right)\left(3x+1+2y\right)\)
e) \(x^2+4x+3\)
\(=\left(x+1\right)\left(x+3\right)\)
d) \(x^2-3x+2\)
\(=\left(x-1\right)\left(x-2\right)\)
Câu 3
cái câu dễ thế mà ko biết đúng là gà