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

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

a) 

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

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

uhm,...excuse me??

2Na + 2H₂O \(\rightarrow\) 2NaOH + H₂ (1)

K₂O + H₂O \(\rightarrow\) 2KOH (2)

Có : m_{dd B} = m_{hh A} + m_H2O - m_H2 = 19,85 + 180,4 - m_H2 = 200

\(\Rightarrow\) m_H2 = 0,25(g)

\(\Rightarrow\) n_H2 = 0,25_/2 = 0,125(mol)

Theo PT(1) \(\Rightarrow\)n_Na = n_NaOH = 2.n_H2 = 2. 0,125 = 0,25(mol)

\(\Rightarrow\left\{{}\begin{matrix}m_{Na}=0,25.23=5,75\left(g\right)\\m_{NaOH}=0,25.40=10\left(g\right)\end{matrix}\right.\)

\(\Rightarrow\) m_K2O = 19,85 - 5,75= 14,1(g)

\(\Rightarrow\) n_K2O = 14,1_/94 = 0,15(mol)

Theo PT(2) \(\Rightarrow\) n_KOH = 2 . n_K2O = 2. 0,15 = 0,3(mol)

\(\Rightarrow\) m_KOH = 0,3 . 56 = 16,8(g)

* C%_{KOH / ddB} = 16,8_/200 . 100% = 8,4%

C%_{NaOH / dd B} = 10_/200 . 100% = 5%

* m_{(KOH+ NaOH)} = 16 ,8 + 10 =\ 26,8(g)

\(\Rightarrow\)m_{H2O / dd B} = 200 - 26,8 = 173,2 (g)

\(\Rightarrow\) V_{H2O / dd B} = m : D = 173,2 : 1 = 173,2 (ml) =0,1732(l)

mà V_{dd B} = V_{H2O / ddB}

=> V_{dd B} =\ 0,1732(l)

Do đó :

C_M của NaOH / dd B = 0,25_/0,1732=1,44(M)

C_M của KOH / dd B = 0,3_/0,1732 = 1,73 ₍M)

gọi x,y là số mol Na và K2O

Theo gt:\(m_A=mNa+mK2O=23x+94y=19,85g\)(1)

Ta có PTHH:2Na+2H2O->2NaOH+H2(1)

........................x......................x.........\(\dfrac{x}{2}\)..(mol)

K2O+H2O->2KOH(2)

.......................y.....................2y.......(mol)

Theo PTHH(1):mH2=\(\dfrac{x}{2}\).2=x (g)

Sau Pưhh,ta có:\(m_B\)=\(m_A\)+mH2O-mH2(bay ra)=19,85+180,4-x=200g

=>x=0,25 mol(2)

Thay (2) vào (1) ta có:\(\begin {cases} x=0,25mol\\ y=0,15mol \end {cases}\)

Theo pt (1);(2):\(\begin {cases} mNaOH=40x=40.0,25=10g\\ mKOH=56y=56.0,15=8,4g \end {cases}\)

*Nồng độ phần trăm:

C%NaOH=(mNaOH:mddB).100%=(10:200).100%=5%

C%KOH=(8,4:200).100%=4,2%

*Nồng độ mol:

ta có:mH2O=180,4g=>VH2O=m:D=180,4:1=180,4 ml=0,1804 lít

Vì VddB=VH2O nên:VddB=0,1804 lít

CM(NaOH)=nNaOH:VddB=0,25:0,1804=1,4M

CM(KOH)=nKOH:VddB=0,15:0,1804=0,83M

a) \(C\%=\dfrac{m_{KCl}}{m_{ddKCl}}.100\%=\dfrac{10}{300}.100\%\approx3,3\%\)

b) Đổi: \(1500ml=1,5l\)

\(C_{MCuSO_4}=\dfrac{n}{V}=\dfrac{3}{1,5}=2M\)

C%NaOH=\(\dfrac{20}{200}100\)=10%

2 

mNaCl= 34g

=>C%NaCl=\(\dfrac{34}{200}.100\)=17%

3

m KOH=20g

=>C%=\(\dfrac{35}{15+200}\)=16,279%

4

C%KCl=\(\dfrac{25}{275}100\)=9,09%