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\(n_{NaOH}=2.0,5=1\left(mol\right)\)
\(n_{KOH}=1.0,5=0,5\left(mol\right)\)
=> Chất tan có trong dung dịch thu được: \(\left\{{}\begin{matrix}NaOH:1\left(mol\right)\\KOH:0,5\left(mol\right)\end{matrix}\right.\)
\(m_{ddNaOH}=1,2.500=600\left(g\right)\)
\(m_{ddKOH}=1,2.500=600\left(g\right)\)
\(\Rightarrow m_{ddsau}=600+600=1200\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaOH}=\dfrac{1.40.100}{12000}=\dfrac{1}{3}\%\\C\%_{KOH}=\dfrac{0,5.56.100}{12000}=\dfrac{7}{30}\%\end{matrix}\right.\)
\(V_{ddsau}=0,5+0,5=1\left(l\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C_{M_{NaOH}}=\dfrac{1}{1}=1\left(M\right)\\C_{M_{KOH}}=\dfrac{0,5}{1}=0,5\left(M\right)\end{matrix}\right.\)
mk làm mà ko chắc đâu nhé,mong mọi người góp ý
áp dụng sơ đồ đường chéo
0,5l dd NaOH 2M------------x-1 ( M)
--------------------------x(M)
0,5l dd KOH 1 M--------------2-x (M)
Ta có x-1= 2-x => x= 1,5 M
mdd NaOH = 500.1,2=600 gam
mdd KOH =500.1,2=600 gam
mNaOH= (0,5.2).40=40 gam ->C%(NaOH)= 40.100%/600=20/3%
mKOH=(0,5.1).56=28 gam -> C%(KOH) =28.100%/600=14/3 %
Áp dụng sơ đồ đường chéo ta tính được
C%=5,666%
\(TC:\)
\(V_1+V_2=2\left(l\right)\)
\(m_{dd_{NaOH\left(3\%\right)}}=1.05V_1\left(g\right)\)
\(m_{NaOH\left(3\%\right)}=1.05V_1\cdot3\%=0.0315V_1\left(g\right)\)
\(m_{dd_{NaOH\left(10\%\right)}}=1.12V_2\left(g\right)\)
\(m_{NaOH\left(10\%\right)}=1.12V_2\cdot10\%=0.112V_2\left(g\right)\)
\(m_{NaOH\left(8\%\right)}=2000\cdot1.1\cdot8\%=176\left(g\right)\)
\(\Leftrightarrow0.0315V_1+0.112V_2=176\left(2\right)\)
\(\left(1\right),\left(2\right):V_1=596\left(ml\right),V_2=1404\left(ml\right)\)
Gọi V dd NaOH 3% = a(lít) ; V dd NaOH 10% = b(lít)
Ta có : a + b = 2(1)
Áp dụng CT : m dd = D.V
m dd NaOH 3% = a.1,05.1000 = 1050a(gam)
m dd NaOH 10% = b.1,12.1000 = 1120b(gam)
m dd NaOH 8% = 2.1,1.1000 = 2200(gam)
Sau khi pha :
m NaOH = 1050a.3% + 1120b.10% = 2200.8%(2)
Từ (1)(2) suy ra a = 0,596(lít) = 596(ml) ; b = 1,404(lít) = 1404(ml)
\(V_{ddNaOH\left(tổng\right)}=400+200=600\left(ml\right)=0,6\left(l\right)\\ n_{NaOH\left(tổng\right)}=0,4.0,5+0,2.1,5=0,5\left(mol\right)\\ C_{MddNaOH\left(cuối\right)}=\dfrac{0,5}{0,6}\approx0,833\left(M\right)\)
\(\left\{{}\begin{matrix}n_{NaOH\left(dd.1M\right)}=0,3\left(mol\right)\\n_{NaOH\left(dd.1,5M\right)}=0,2.1,5=0,3\left(mol\right)\end{matrix}\right.\)
\(n_{NaOH\left(dd.sau\right)}=n_{NaOH\left(dd.1M\right)}+n_{NaOH\left(dd.1,5M\right)}=0,3+0,3=0,6\left(mol\right)\)
\(V_{dd\left(sau\right)}=300+200=500\left(ml\right)=0,5\left(l\right)\)
\(\Rightarrow CM_{dd\left(sau\right)}=\frac{0,6}{0,6}=1,2M\)
\(\left\{{}\begin{matrix}m_{dd.sau}=500.1,05=525\left(g\right)\\m_{NaOH}=06.40=24\left(g\right)\end{matrix}\right.\)
\(\Rightarrow C\%_{Dd\left(spu\right)}=\frac{24}{525}.100\%=4,57\%\)
nNaOH trong dd NaOH 1M=0,3(mol)
nNaOH trong dd NaOH 1,5M=0,3(mol)
CM=\(\dfrac{0,3+0,3}{0,5}=1,2M\)
C%=\(\dfrac{40.1,2}{10.0,5}=9,6\%\)
nNaOH=0,3.1+0,2.1,5=0,6(mol)
CM= 0,6/(0,3+0,2)=1,2M
m=1,05.(300+200)=525(g)
mNaOH=0,6.40=24(g)
C%=24/525.100%=4,57%
Tính C% của dung dịch thu được:
Ta có: md d NaOH(1)=V.D=500.1,2=600(g)
Vd d NaOH(1)=500ml=0,5 (lít)
=> nNaOH(1)=CM.V=2.0,5=1 (mol)
=> mNaOH(1)=nNaOH.M=1.40=40(gam)
Ta có: md d NaOH(2)=V.D=300.1,1=330(g)
Vd d NaOH(2)=300ml=0,3 (lít)
=> nNaOH(2)=CM.V=0,5.0,3=0,15(mol)
=> mNaOH(2)=n.M=0,15.40=6(gam)
=> mNaOH mới=mNaOH(1) + mNaOH(2)=40+6=46(gam)
md d NaOH mới=md d NaOH(1) + md d NaOH(2)=600+330=930(gam)
=> \(C\%_{ddsauphanung}=\dfrac{m_{NaOHmới}.100\%}{m_{ddNaOHmoi}}=\dfrac{46.100}{930}\approx4,95\left(\%\right)\)
Tính CM của dung dịch thu được :
Ta có: Vd d NaOH mới=Vd d NaOH(1) + Vd d NaOH(2)=0,5+0,3=0,8(lít)
nd d NaOH mới= n d d NaOH(1) + n d d NaOH(2)= 1 + 0,15=1,15(mol)
=> \(C_M=\dfrac{n}{V}=\dfrac{1,15}{0,8}\approx1,44\left(M\right)\)