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\(m_{HCl3\%}=\dfrac{500.3}{100}=15\left(g\right)\)
\(m_{HCl10\%}=\dfrac{300.10}{100}=30\left(g\right)\)
\(C\%=\dfrac{15+30}{500+300}.100\%=5,625\%\)
Vậy ta được dd mới có nồng độ 5,625%
\(m_{HCl\left(1\right)}=600.2,5\%=15g\\ m_{HCl\left(2\right)}=400.15\%=60g\\ C\%_{HCl.sau}=\dfrac{75}{1000}\cdot100\%=7,5\%\)
1) Ta có: \(m_{H_2SO_4}=200\cdot15\%+300\cdot25\%=105\left(g\right)\)
\(\Rightarrow C\%_{H_2SO_4}=\dfrac{105}{200+300}\cdot100\%=21\%\)
2) Ta có: \(\left\{{}\begin{matrix}n_{H_2SO_4}=\dfrac{105}{98}=\dfrac{15}{14}\left(mol\right)\\V_{ddH_2SO_4}=\dfrac{500}{1,25}=400\left(ml\right)\end{matrix}\right.\) \(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{\dfrac{15}{14}}{0,4}\approx2,68\left(M\right)\)
a)m dd sau=100gam
mNaCl không đổi=80.15%=12 gam
C% dd NaCl sau=12/100.100%=12%
b)mdd sau=200+300=500 gam
Tổng mNaCl sau khi trộn=200.20%+300.5%=55 gam
C% dd NaCl sau=55/500.100%=11%
c) mdd sau=150 gam
mNaOH trg dd 10%=5 gam
mNaOH trong dd sau khi trộn=150.7,5%=11,25 gam
=>mNaOH trong dd a%=11,25-5=6,25 gam
=>C%=a%=6,25/100.100%=6,25% => a=6,25
\(\text{1)}m_{KOH}=40.35\%=14\left(g\right)\\ \rightarrow n_{KOH}=\dfrac{14}{56}=0,25\left(mol\right)\\ PTHH:KOH+HCl\rightarrow KCl+H_2O\\ \text{Theo pthh}:n_{HCl}=n_{KOH}=0,25\left(mol\right)\\ \rightarrow V_{ddHCl}=0,25.0,5=0,125\left(l\right)\)
\(\text{2)}n_{Al}=\dfrac{4,05}{27}=0,15\left(mol\right)\\ n_{H_2SO_4}=200.14,7\%=29,4\left(g\right)\\ \rightarrow n_{H_2SO_4}=\dfrac{29,4}{98}=0,3\\ \text{PTHH}:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\\ \text{LTL}:\dfrac{0,15}{2}< \dfrac{0,3}{3}\rightarrow H_2SO_4\text{ dư}\)
\(\text{Theo pthh}:\left\{{}\begin{matrix}n_{H_2SO_4\left(pư\right)}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,15=0,225\left(mol\right)\\n_{H_2}=n_{H_2SO_4\left(pư\right)}=0,225\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}.0,15=0,075\left(mol\right)\end{matrix}\right.\\ \rightarrow m_{dd\left(\text{sau phản ứng}\right)}=200+4,05-0,3.2=203,45\left(g\right)\)
\(\rightarrow\left\{{}\begin{matrix}C\%_{H_2SO_4\text{ dư}}=\dfrac{\left(0,3-0,225\right).98}{203,45}=3,61\%\\C\%_{Al_2\left(SO_4\right)_3}=\dfrac{342.0,075}{203,45}=12,61\%\end{matrix}\right.\)
\(C\%_{H_2SO_4}=\dfrac{50\cdot16\%+40\cdot60\%}{50+40}=35.55\%\)
mddH2SO4(sau)=50+40=90(g)
mH2SO4(tổng)= 50.46%+40.60%=47(g)
=>C%ddH2SO4(sau)= (47/90).100=52,222%
1)
$m_{dd} = 50 + 30 = 80(gam)$
$m_{KOH} = 50.20\% + 30.15\% = 14,5(gam)$
$C\% = \dfrac{14,5}{80}.100\% = 18,125\%$
2)
$m_{dd} = 200 + 300 = 500(gam)$
$m_{NaCl} = 200.20\% + 300.5\% = 55(gam)$
$C\% = \dfrac{55}{500}.100\% = 11\%$
\(m_{dd}=50+35=85\left(g\right)\\ m_{ct}=\left(\dfrac{50.35}{100}\right)+\left(\dfrac{35.8}{100}\right)=20,3\left(g\right)\\ C\%=\dfrac{20,3}{85}.100\%=23,88\left(g\right)\)