Câu 11
K
Khách
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Những câu hỏi liên quan
21 tháng 5 2021
Câu 1:
\(-12+\left(16-11\right).4=-12+5.4=-12+20=8\)
Câu 2:
\(2\dfrac{3}{7}+1\dfrac{4}{7}=\dfrac{17}{7}+\dfrac{11}{7}=\dfrac{17+11}{7}=\dfrac{28}{7}=4\)
Câu 3:
\(\dfrac{-2}{3}.\dfrac{4}{5}+\dfrac{1}{5}:\dfrac{9}{11}=\dfrac{-8}{15}+\dfrac{11}{45}=\dfrac{-13}{45}\)
H
21 tháng 5 2021
Câu 1:
\(-12+\left(16-11\right).4=-12+5.4=-12+20=8\)
Câu 2:
\(2\dfrac{3}{7}+1\dfrac{4}{7}=\dfrac{17}{7}+\dfrac{11}{7}=\dfrac{28}{7}=4\)
Câu 3:
\(\dfrac{-2}{3}.\dfrac{4}{5}+\dfrac{1}{5}:\dfrac{9}{11}=\dfrac{-8}{15}+\dfrac{1}{5}.\dfrac{11}{9}=\dfrac{-8}{15}+\dfrac{11}{45}=\dfrac{-13}{45}\)
19 tháng 11 2023
Câu 10: Đúng
Câu 11: Cấm đi ngược chiều
Câu 14; Đúng
21 tháng 5 2021
Câu 1:
\(\dfrac{2}{5}-\dfrac{1}{4}+\dfrac{3}{10}=\dfrac{8}{20}-\dfrac{5}{20}+\dfrac{6}{20}=\dfrac{8-5+6}{20}=\dfrac{9}{20}\)
Câu 2:
\(\dfrac{-2}{5}:\left(1-\dfrac{1}{10}\right)=\dfrac{-2}{5}:\dfrac{9}{10}=\dfrac{-2}{5}.\dfrac{10}{9}=\dfrac{-2.10}{5.9}=\dfrac{-20}{45}=\dfrac{-4}{9}\)
Câu 3:
\(\dfrac{7}{8}.\dfrac{4}{9}+\dfrac{1}{14}:\dfrac{5}{14}=\dfrac{7}{18}+\dfrac{1}{5}=\dfrac{53}{90}\)
Câu 4:
\(\dfrac{2}{7}.\dfrac{3}{11}+\dfrac{2}{7}.\dfrac{8}{11}\)
\(=\dfrac{2}{7}.\left(\dfrac{3}{11}+\dfrac{8}{11}\right)\)
\(=\dfrac{2}{7}.1\)
\(=\dfrac{2}{7}\)
S
21 tháng 5 2021
Câu 1
\(\dfrac{2}{5}\)-\(\dfrac{1}{4}\)+\(\dfrac{3}{10}\)= \(\dfrac{8}{20}\)-\(\dfrac{5}{20}\)+\(\dfrac{6}{20}\)=\(\dfrac{3}{20}\)+\(\dfrac{6}{20}\)=\(\dfrac{9}{20}\)
Câu 2
-\(\dfrac{2}{5}\):(1-\(\dfrac{1}{10}\))= -\(\dfrac{2}{5}\):\(\dfrac{9}{10}\)=-\(\dfrac{2}{5}\).\(\dfrac{10}{9}\)=-\(\dfrac{4}{9}\)
Câu 3
\(\dfrac{7}{8}.\dfrac{4}{9}+\dfrac{1}{14}:\dfrac{5}{14}\)= \(\dfrac{7}{8}.\dfrac{4}{9}+\dfrac{1}{14}.\dfrac{14}{5}\)=\(\dfrac{7.4}{4.2.9}+\dfrac{1.14}{14.5}\)=\(\dfrac{7}{18}+\dfrac{1}{5}\)=\(\dfrac{35}{90}+\dfrac{18}{90}\)=\(\dfrac{53}{90}\)
Câu 4
\(\dfrac{2}{7}.\dfrac{3}{11}+\dfrac{2}{7}.\dfrac{8}{11}\)=\(\dfrac{2}{7}.\left(\dfrac{3}{11}+\dfrac{8}{11}\right)\)=\(\dfrac{2}{7}.1\)=\(\dfrac{2}{7}\)
LM
5
TP
27 tháng 12 2021
\(7-B\\ 8-A\\ 9-C.Tacó:n_M=n_{MCl}\\ \Rightarrow\dfrac{4,6}{M}=\dfrac{11,7}{M+35,5}\\ \Rightarrow M=23\left(Na\right)\\ 10-A.2NaOH+Cl_2\rightarrow NaCl+NaClO+H_2O\\ 11-D\\ 12-B\\ 13-B\\ 14-D.BTNT\left(S\right):n_{H_2SO_4}=n_{SO_3}=\dfrac{16}{80}=0,2\left(mol\right)\\ CM_{H_2SO_4}=\dfrac{0,2}{0,25}=0,8M\)
\(n_{HCl}=1,4.0,5=0,7\left(mol\right)\\ n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ Vì:\dfrac{0,7}{2}>\dfrac{0,2}{1}\Rightarrow HCldư\\ \Rightarrow ddsau:\left\{{}\begin{matrix}HCl\left(dư\right)\\CuCl_2\end{matrix}\right.\\ V_{ddsau}=V_{ddHCl}=500\left(ml\right)=0,5\left(l\right)\\ n_{CuCl_2}=n_{CuO}=0,2\left(mol\right)\\ n_{HCl\left(dư\right)}=0,7-0,2.2=0,3\left(mol\right)\\ m_{CuCl_2}=135.0,2=27\left(g\right)\\ m_{HCl\left(dư\right)}=0,3.36,5=10,95\left(g\right)\\ C_{MddHCl}=\dfrac{0,3}{0,5}=0,6\left(M\right)\\ C_{MddCuCl_2}=\dfrac{0,2}{0,5}=0,4\left(M\right)\)