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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, \(m_{NaOH}=10\%.200+5\%.300=35\left(g\right)\)
\(\rightarrow C\%_{NaOH}=\dfrac{35}{200+300}=7\%\)
Na2O=0,5 mol
Na2O+H2O->2NaOH
0,5-----------------1 mol
ta có m NaOH=1.40+40=80g
=>C%=\(\dfrac{80}{431}100=18,561\%\)
a. Khối lượng chất tan: 10g
khối lượng dung dịch : 100g
b.
Khối lượng chất tan: 6,2 g
khối lượng dung dịch :111,6g
c.
Khối lượng chất tan: 4,6g
khối lượng dung dịch :154,6g
d.
Khối lượng chất tan: 6,5g
khối lượng dung dịch :206,5g
\(a,m_{ct}=10\left(g\right)\\ m_{dd}=10+90=100\left(g\right)\)
b, \(n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)
PTHH: Na2O + H2O ---> 2NaOH
0,1 0,2
\(m_{ct}=0,2.40=8\left(g\right)\\ m_{dd}=6,2+105,4=111,6\left(g\right)\)
c, \(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)
PTHH: 2Na + 2H2O ---> 2NaOH + H2
0,2 0,2 0,1
mct = 0,2.40 = 8 (g)
\(m_{dd}=4,6+150-0,1.2=154,4\left(g\right)\)
d, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PTHH: Zn + 2HCl ---> ZnCl2 + H2
0,1 0,1 0,1
mct = 0,1.136 = 13,6 (g)
mdd = 200 + 6,5 - 0,2 = 206,3 (g)
\(m_{NaOH\left(bđ\right)}=\dfrac{90,7.8}{100}=7,256\left(g\right)\)
\(n_{Na_2O}=\dfrac{a}{62}\left(mol\right)\)
PTHH: Na2O + H2O --> 2NaOH
\(\dfrac{a}{62}\)------------->\(\dfrac{a}{31}\)
=> \(m_{NaOH\left(sau.pư\right)}=\dfrac{a}{31}.40+7,256\left(g\right)\)
mdd sau pư = a + 90,7 (g)
=> \(C\%_{dd.sau.pư}=\dfrac{\dfrac{40}{31}a+7,256}{a+90,7}.100\%=12\%\)
=> a = 3,1 (g)
\(a.n_{Na_2O}=\dfrac{0,62}{62}=0,01\left(mol\right)\\ Na_2O+H_2O\rightarrow2NaOH\\ n_{NaOH}=2n_{Na_2O}=0,02\left(mol\right)\\ C\%_{NaOH}=\dfrac{0,02.40}{0,62+3,38}.100=20\%\\ b.HCl+NaOH\rightarrow NaCl+H_2O\\ n_{HCl}=n_{NaOH}=0,02\left(mol\right)\\ \Rightarrow V_{HCl}=\dfrac{0,01}{1}=0,01\left(l\right)\)
a, \(Na_2O+H_2O\rightarrow2NaOH\)
Ta có: \(n_{Na_2O}=\dfrac{3,1}{62}=0,05\left(mol\right)\)
Theo PT: \(n_{NaOH}=2n_{Na_2O}=0,1\left(mol\right)\)
m dd sau pư = 3,1 + 50 = 53,1 (g)
\(\Rightarrow C\%_{NaOH}=\dfrac{0,1.40}{53,1}.100\%\approx7,53\%\)
b, \(2Na+2H_2O\rightarrow2NaOH+H_2\)
Ta có: \(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)
Theo PT: \(\left\{{}\begin{matrix}n_{NaOH}=n_{Na}=0,2\left(mol\right)\\n_{H_2}=\dfrac{1}{2}n_{Na}=0,1\left(mol\right)\end{matrix}\right.\)
Ta có: m dd sau pư = 4,6 + 95,6 - 0,1.2 = 100 (g)
\(\Rightarrow C\%_{NaOH}=\dfrac{0,2.40}{100}.100\%=8\%\)
mNACl=200.2%+300.5%=19(g)
C%=\(\dfrac{19}{200+300}.100\%=0,18\%\)
\(m_{ddNaOH\left(sau\right)}=200+300=500\left(g\right)\\ m_{NaOH}=200.2\%+300.5\%=19\left(g\right)\\ C\%_{ddNaOH}=\dfrac{19}{500}.100=3,8\%\)
\(a,C\%_{CuSO_4}=\dfrac{5}{200+5}.100\%=2,43\%\\ b,C\%_{NaOH}=\dfrac{0,2.40}{200}.100\%=4\%\\ c,n_{NH_3}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ C\%_{NH_3}=\dfrac{0,3.17}{200+0,3.17}.100\%=2,5\%\\ d,n_{KCl}=\dfrac{9.10^{22}}{6.10^{23}}=0,15\left(mol\right)\\ C\%_{KCl}=\dfrac{0,15.74,5}{200}=5,5875\%\)
Khối lượng của dung dịch là:
6,2 + 200 = 206,2 (g )
C%= mchất tan : mdungdịch . 100% = 6,2: 206,2. 100= 3,006 %