\(\dfrac{m_1}{m_2}\) = \(\dfrac{C_2-C}{C-C_1}\)<=> \(\dfrac{50}{50}\) = \(\dfrac{20-C}{C-5}\)

=> C = 12,5 %

a) \(m_{NaCl}=80\times15\%=12\left(g\right)\)

\(m_{ddNaCl}mới=20+80=100\left(g\right)\)

\(\Rightarrow C\%_{NaCl}mới=\frac{12}{100}\times100\%=12\%\)

b) \(m_{NaCl.20\%}=200\times20\%=40\left(g\right)\)

\(m_{NaCl.5\%}=300\times5\%=15\left(g\right)\)

\(\Rightarrow m_{NaCl}mới=40+15=55\left(g\right)\)

\(m_{ddNaCl}mới=200+300=500\left(g\right)\)

\(\Rightarrow C\%_{NaCl}mới=\frac{55}{500}\times100\%=11\%\)

c) \(m_{H_2SO_4.10\%}=100\times10\%=10\left(g\right)\)

\(m_{H_2SO_4.25\%}=150\times25\%=37,5\left(g\right)\)

\(\Rightarrow m_{H_2SO_4}mới=10+37,5=47,5\left(g\right)\)

\(\Rightarrow m_{ddH_2SO_4}mới=100+150=250\left(g\right)\)

\(\Rightarrow C\%_{H_2SO_4}mới=\frac{47,5}{250}\times100\%=19\%\)

Áp dụng quy tắc đường chéo ta có:

a) \(D_1=20g\)

\(D_2=80g\)

\(\frac{D_1}{D_2}=\frac{20}{80}=\frac{15-C\%}{C\%}\rightarrow C\%=12\%\)

b) \(D_1=200\left(g\right)\)

\(D_2=300\left(g\right)\)

\(\frac{D_1}{D_2}=\frac{200}{300}=\frac{C\%-5}{20-C\%}\rightarrow C\%=11\%\)

c) \(D_1=100\left(g\right)\)

\(D_2=150\left(g\right)\)

\(\frac{D_1}{D_2}=\frac{100}{150}=\frac{25-C\%}{C\%-10}\rightarrow C\%=19\%\)

Sửa lại đề câu 1 tớ hộ hết

a, m_{ct ^{= \(\dfrac{15\cdot80}{100}\)}}= 12 (g)

C% = \(\dfrac{12}{80+20}\)* 100 = 12%

b, áp dụng quy tắc đương chéo:

m1= 200g có C%=20%↓ C-5

C%➚

m2=300g có c% =5%➚ 20-C

ta có:

\(\dfrac{200}{300}\)=\(\dfrac{2}{3}\)=\(\dfrac{C-5}{20-C}\)⇒2.(20-C)=3.(C-5)

Giải pt ta được C=11%

a, mct = $\frac{15\cdot 80}{100}$= 12 (g)

C% = $\frac{12}{80+20}$* 100 = 12%

b, áp dụng quy tắc đương chéo:

m1= 200g có C%=20%↓ C-5

C%➚

m2=300g có c% =5%➚ 20-C

ta có:

$\frac{200}{300}$=$\frac{2}{3}$=$\frac{C-5}{20-C}$⇒2.(20-C)=3.(C-5)

Giải pt ta được C=11%

Gọi C_%(Y)=x (%)

Áp dụng quy tắc đường chéo ta có

200g dd NaOH 10%                           20 - x

                                                  x

300g dd NaOH 20%                              x-10

=>\(\frac{200}{300}\)= \(\frac{20-x}{x-10}\)

=> x=16 (%)

Vậy nồng đọ % của dd Y là 16%

\(n_{CuSO4}=\dfrac{16\%.50}{100\%.160}=0,05\left(mol\right)\)

Pt : \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

       0,05-------->0,1---------->0,05--------->0,05

a) \(C\%_{ddNaOH}=\dfrac{0,1.40}{250}.100\%=1,6\%\)

b) \(m_{ddspu}=50+250-0,05.98=295,1\left(g\right)\)

\(C\%_{Na2SO4}=\dfrac{0,05.142}{295,1}.100\%=2,41\%\)

\(n_{CuSO_4}=\dfrac{m_{dd}\cdot C\%}{100\cdot M}=\dfrac{50\cdot16\%}{100\cdot\left(64+32+16\cdot4\right)}=0,05\left(mol\right)\)

\(PTHH:CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

                     1                  2                    1                    1

                   0,05              0,1                 0,05               0,05 (mol)

\(a)C\%_{NaOH}=\dfrac{n\cdot100\cdot M}{m_{dd}}=\dfrac{01\cdot100\cdot\left(23+16+1\right)}{250}=1,6\%\)

\(b)m_{dd-sau-pư}=m_{dd_đ}+m_{ct_đ}-m\downarrow-m\uparrow\)

\(=50+250-\left(0,05\cdot23+32+16\cdot4\right)=294,05\left(g\right)\)

\(C\%_{Na_2SO_4}=\dfrac{n\cdot100\cdot M}{m_{dd}}=\dfrac{0,05\cdot100\cdot\left(23\cdot2+32+16\cdot4\right)}{294,05}\approx2,41\%.\)

nCuSO4​=100%.16016%.50​=0,05(mol)

Pt : ����4+2����→��(��)2+��2��4CuSO4​+2NaOH→Cu(OH)2​+Na2​SO4​

       0,05-------->0,1---------->0,05--------->0,05

a) �%������=0,1.40250.100%=1,6%C%ddNaOH​=2500,1.40​.100%=1,6%

b) ������=50+250−0,05.98=295,1(�)mddspu​=50+250−0,05.98=295,1(g)

�%��2��4=0,05.142295,1.100%=2,41%C%Na2SO4​=295,10,05.142​.100%=2,41%

a)m_H2SO4=\(\dfrac{200.7,3\text{%}}{100\%}\)=14,6g

n_HCl=\(\dfrac{14,6}{36,5}\)=0,4(mol)

PTHH:

NaOH+ HCl→ NaCl+ H₂O

1            1            1         1

0,4        0,4         0,4     (mol)

⇒m_NaOH=0,4.40=16(g)

Nồng độ % của dd NaOH cần dùng là:

C_%NaOH=\(\dfrac{16}{200}\)   .100%=8%

b)Ta có:m_{dd spứ}=m_{dd trc pứ}=400g

m_NaCl=0,4.58,5=23,4g

Nồng độ % dd muối tạo thành sau pứ là:

C_%dd NaCl=\(\dfrac{23,4}{400}\)   .100%=5,85%        

\(V_{NaClsau}=3+4=7\left(l\right)\)

\(n_{NaCl0,5M}=0,5.3=1,5\left(mol\right)\)

\(n_{NaCl1,5M}=1,5.4=6\left(mol\right)\)

\(C_{M_{NaClsau}}=\dfrac{n}{V}=\dfrac{1,5+6}{7}=\dfrac{15}{14}M\)

a, \(D=\dfrac{m}{V}\Rightarrow m_{ddNa_2SO_4}=1,12.2.1000=2240\left(g\right)\)

b, PT: \(BaCl_2+Na_2SO_4\rightarrow2NaCl+BaSO_{4\downarrow}\)

Ta có: \(m_{BaCl_2}=416.20\%=83,2\left(g\right)\Rightarrow n_{BaCl_2}=\dfrac{83,2}{208}=0,4\left(mol\right)\)

Theo PT: \(n_{BaSO_4}=n_{BaCl_2}=0,4\left(mol\right)\)

\(n_{NaCl}=2n_{BaCl_2}=0,8\left(mol\right)\)

Có: m _{dd sau pư} = 416 + 2240 - 0,4.233 = 2562,8 (g)

\(\Rightarrow C\%_{NaCl}=\dfrac{0,8.58,5}{2562,8}.100\%\approx1,83\%\)

\(n_{BaCl2}=\dfrac{20\%.416}{100\%.208}=0,4\left(mol\right)\)

 Pt : \(BaCl_2+Na_2SO_4\rightarrow BaSO_4\downarrow+2NaCl\)

            0,4                              0,4           0,8

a) \(m_{ddNa2SO4}=D.V=1,12.2000=2240\left(g\right)\)

b) \(m_{ddspu}=416+2240-\left(0,4.233\right)=2562,8\left(g\right)\)

\(C\%=\dfrac{\left(0,8.58,5\right)}{2562,8}.100\%=1,83\%\)

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