\(a,m_{KOH}=\dfrac{28.10}{100}=2,8\left(g\right)\\ \rightarrow n_{KOH}=\dfrac{2,8}{56}=0,05\left(mol\right)\\ b,C\%=\dfrac{36}{144+36}.100\%=20\%\\ c, n_{NaOH}=\dfrac{0,8}{40}=0,02\left(mol\right)\\ \rightarrow C_{M\left(NaOH\right)}=\dfrac{0,02}{0,08}=0,25M\)

\(a,m_{KOH}=\dfrac{28.10}{100}=2,8\left(g\right)\\ n_{KOH}=\dfrac{2,8}{56}=0,05\left(mol\right)\\ C\%=\dfrac{36}{36+144}.100\%=20\%\\ C_M=\dfrac{0,8}{0,08}=10M\)

a)\(n_{NaCl}=\dfrac{11,7}{58,5}=0,2mol\)

\(C_{M_{NaCl}}=\dfrac{n_{NaCl}}{V_{NaCl}}=\dfrac{0,2}{2}=0,1M\)

b)\(n_{KOH}=\dfrac{3,36}{56}=0,06mol\)

\(C_{M_{KOH}}=\dfrac{n_{KOH}}{V_{KOH}}=\dfrac{0,06}{0,3}=0,2M\)

a) 

\(C\%_{dd.KOH}=\dfrac{7,5}{7,5+42,5}.100\%=15\%\)

b) \(n_{HNO_3}=\dfrac{1,26}{63}=0,02\left(mol\right)\Rightarrow C_{M\left(dd.HNO_3\right)}=\dfrac{0,02}{0,016}=1,25M\)

Ta có: n KOH   =   8 , 4 / 56   =   0 , 15   ( mol )

  → C M ( KOH ) =   0 , 15 / 0 , 5   =   0 , 3 M .

Câu 1  :

a) n Na2O = 3,1/62 = 0,05(mol)

$Na_2O + H_2O \to 2NaOH$

Theo PTHH : n NaOH = 2n Na2O = 0,1(mol)

=> CM NaOH = 0,1/2 = 0,05M

Câu 2 : 

Coi n KOH = 1(mol)

=> V dd KOH = 1/2 = 0,5(lít) = 500(ml)

=> mdd KOH = D.V = 500.1,43 = 715(gam)

=> C% KOH = 1.56/715  .100% = 7,83%

1. Ta có : \(n_{Na_2O}=\dfrac{m}{M}=0,05mol\)

\(PTHH:Na_2O+H_2O\rightarrow2NaOH\)

Theo PTHH: \(n_{NaOH}=2n_{Na_2O}=0,1mol\)

\(\Rightarrow C_{MNaOH}=\dfrac{n}{V}=0,05M\)

2. - Gọi số lít KOH là a lít

\(\Rightarrow m_{dd}=D.V=1430a\left(g\right)\)

Mà \(n_{KOH}=C_M.V=2amol\)

\(\Rightarrow m_{KOH}=n.M=112a\left(g\right)\)

\(\Rightarrow C\%=\dfrac{m}{m_{dd}}.100\%=\dfrac{112a}{1430a}.100\%=~7,83\%\)

a) mKOH(A)=150.5%=7,5 gam
Gọi mdd KOH 12% thêm=a gam
=>mKOH thêm=0,12a gam
tổng mKOH=0,12a+7,5 gam
mdd KOH=a+150 gam
C%dd KOH sau=(0,12a+7,5)/(a+150).100%=10%
=>a=375 gam
b)Gọi mKOH thêm=b gam
tổng mKOH sau=b+7,5 gam
mdd KOH sau=b+150 gam
C% dd KOH sau=10%
=>0,1(b+150)=b+7,5
=>b=8,3333 gam
c)Làm bay hơi=> Gọi mH2O tách ra=c gam
mdd sau=150-c gam
mKOH sau=7,5 gam
C% dd KOH sau=7,5/(150-c).100%=10%
=>c=75 gam
=>mdd KOH 10% sau=75 gam

 

thanks bạn

a) \(m_{NaOH}=60.5\%=3\left(g\right)\)

\(m_{dd}=40+60=100\left(g\right)\)

\(\Rightarrow C\%\left(NaOH\right)=\dfrac{3}{100}.100\%=3\%\)

b) \(m_{dd}=100+50=150\left(g\right)\)

\(m_{KOH}=100.20\%+50.15\%=27,5\left(g\right)\)

\(\Rightarrow C\%\left(KOH\right)=\dfrac{27,5}{150}.100\%=18,33\%\)

a, m_ddsau = 40 + 60 = 100 ( g )

m_NaOH = 3 ( g )

=> \(C\%=\dfrac{3}{100}.100=3\%\)

b, - Gọi nồng độ dd thu được là X % .

- Áp đụng pp đường chéo ta có :

\(\Rightarrow\dfrac{100}{50}=2=\dfrac{X-15}{20-X}\)

=> X = 55/3 %

uhm,...excuse me??