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a) \(m_{dd}=\dfrac{200.100}{10}=2000\left(g\right)\)
b) mH2O = 2000 - 200 = 1800 (g)
a, \(n_{NaOH}=0,2.1=0,2\left(mol\right)\)
\(m_{NaOH}=0,2.40=8\left(g\right)\)
b, \(n_{H_2SO_4}=2.0,1=0,2\left(mol\right)\)
\(c,C\%=\dfrac{6}{200}.100\%=3\%\)
\(m_{NaCl}=\dfrac{200.8}{100}=16\left(g\right)\)
a.\(n_{NaOH}=\dfrac{8}{40}=0,2mol\)
\(V_{dd}=\dfrac{120}{1,2}=100ml=0,1l\)
\(C_{M_{NaOH}}=\dfrac{0,2}{0,1}=2M\)
b.\(n_{NaOH}=\dfrac{21,6}{40}=0,54mol\)
\(V_{dd}=\dfrac{180}{1,2}=150ml=0,15l\)
\(C_{M_{NaOH}}=\dfrac{0,54}{0,15}=3,6M\)
Theo đề: mddNaOH= 650.1,114= 724,1 (g)
Gọi khối lượng Na2O cần dùng là a gam (a>0)
Ta có quy tắc đường chéo:
=> \(\dfrac{a}{724,1}=\dfrac{30}{60}=\dfrac{1}{2}\)
=> a= 362,05 (g)
Vậy cần dùng 362,05 gam Na2O
a)\(C\%=\dfrac{m_{ct}}{m_{dd}}\cdot100\%=\dfrac{5}{5+45}\cdot100\%=10\%\)
b)\(C\%=\dfrac{m_{ct}}{m_{dd}}\cdot100\%=\dfrac{5,6}{5,6+94,4}\cdot100\%=5,6\%\)
c)\(m_{ctNaOH}=\dfrac{200\cdot10\%}{100\%}=20g\)
\(m_{ctNaOH}=\dfrac{300\cdot5\%}{100\%}=15g\)
\(C\%=\dfrac{m_{ct}}{m_{dd}}\cdot100\%=\dfrac{20+15}{200+300}\cdot100\%=7\%\)
\(a,C\%_{NaOH}=\dfrac{5}{5+45}=10\%\)
b, \(n_{CaO}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PTHH: CaO + H2O ---> Ca(OH)2
0,1 ---------------> 0,1
\(\rightarrow C\%_{Ca\left(OH\right)_2}=\dfrac{74.0,1}{5,6+94,4}=37\%\)
c, \(m_{NaOH}=10\%.200+5\%.300=35\left(g\right)\)
\(\rightarrow C\%_{NaOH}=\dfrac{35}{200+300}=7\%\)
\(a.m_{ddNaCl}=\dfrac{15}{5}\cdot100=300g\\ b.m_{nước}+m_{muối}=m_{dd,muối}\\ \Rightarrow m_{nước}=m_{dd,muối}-m_{muối}\\ =300-15\\ =285g\)
\(a,C_{M\left(NaOH\right)}=\dfrac{0,3}{0,5}=0,6M\\ b,n_{NaOH}=\dfrac{24}{40}=0,6\left(mol\right)\\ C_{M\left(NaOH\right)}=\dfrac{0,6}{0,4}=1,5M\)
\(m_{NaOH}=a\left(g\right)\)
\(m_{dd_{NaOH}}=m_{NaOH}+m_{H_2O}=a+200\left(g\right)\)
\(C\%_{NaOH}=\dfrac{m_{NaOH}}{m_{dd_{NaOH}}}\cdot100\%=8\%\)
\(\Leftrightarrow\dfrac{a}{200+a}\cdot100\%=8\%\)
\(\Leftrightarrow200+a=12.5a\)
\(\Leftrightarrow a=17.4\left(g\right)\)
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