lớp 1 đã học cái này rùi cơ à bạn

n_Fe=m:M=2,8:56=0,05mol

n_CuO=m:M=4:80=0,05mol

Fe+H₂SO₄--->FeSO₄+H₂

_{0,05mol. --------->0,05mol}

H₂+CuO---->Cu +H₂O

0,05mol--->0,05mol

V_H2=n. 22,4=0,05.22,4=1,12lit

m_Cu=n.M=0,05.64=3,2g

ta có:

\(\frac{1}{4^2}+\frac{1}{6^2}+..+\frac{1}{\left(2n\right)^2}=\frac{1}{\left(2.2\right)^2}+\frac{1}{\left(2.3\right)^2}+...+\frac{1}{\left(2n\right)^2}=\frac{1}{2^2.2^2}+\frac{1}{2^2.3^2}+...+\frac{1}{2^2.n^2}\)

\(=\frac{1}{2^2}.\frac{1}{2^2}+\frac{1}{2^2}.\frac{1}{3^2}+..+\frac{1}{2^2}.\frac{1}{n^2}=\frac{1}{2^2}.\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)=\frac{1}{4}.\left(\frac{1}{2^2}+\frac{1}{3^2}+..+\frac{1}{n^2}\right)\)

mà 1/2^2+1/3^2+..+1/n^2 < 1(cái này bn tự c/nm đc chứ?)

=>\(\frac{1}{4}.\left(\frac{1}{2^2}+\frac{1}{3^2}+..+\frac{1}{n^2}\right)<\frac{1}{4}\left(đpcm\right)\)

tôi cho Hoàng Phúc

C_M=n:V

=>n=C_M.V

=>V=n:C_M

đừng nghe, bạn ấy làm thế để các cậu xóa tab này đấy

1.(3x-2)⁵=-243=(-3)⁵

=>3x-2=-3

=>3x=-3+2=-1

=>x=-1/3

2.1/9.27^x=3^x

=>\(\frac{1}{9}=3^x:27^x=\frac{3^x}{27^x}=\left(\frac{3}{27}\right)^x=\left(\frac{1}{9}\right)^x\Rightarrow x=1\)

3.1/3².3⁴.3ⁿ=3⁷

=>3^-2.3⁴.3ⁿ=3⁷

=>3^-2+4+n=3⁷

=>(-2)+4+n=7

=>2+n=7=>n=5

Vậy n=5

x*3+27=60

X*3=60-27

X*3=33

X=33:3

X=11

FeO+H₂SO₄--->H₂O+FeSO₄

_{0,15mol--->0,15mol--->0,15mol---->0,15mol}

n_FeO=m:M=10,8:72=0,15mol

m_FeSO4=n.M=0,15.152=22,8g

V_dd=m_dm+m_ct=0,15.(18+152)=25,5g

C%=m_ct:m_dd.100%=22,8:25,5.100%≈89,41%

V_H2SO4=n.22,4=0,15.22,4=3,36lit