Em đăng vào môn Hoá nhé!

Nồng độ phần trăm:

\(C\%=\dfrac{8,4}{508,4}.100\%\approx1,7\%\)

\(CaO+H_2O->Ca\left(OH\right)_2\)

0,2..........................0,2

n CaO = \(\dfrac{11,2}{40+16}=0,2mol\)

m Ca(OH)2 = \(0,2.\left(40+16.2+1.2\right)=14,8g\)

C% Ca(OH)2 = \(\dfrac{14,8}{500}.100=2,96\%\)

a, \(C\%_{KCl}=\dfrac{20}{20+60}.100\%=25\%\)

b, \(C\%=\dfrac{40}{40+150}.100\%\approx21,05\%\)

c, \(C\%_{NaOH}=\dfrac{60}{60+240}.100\%=20\%\)

d, \(C\%_{NaNO_3}=\dfrac{30}{30+90}.100\%=25\%\)

e, \(m_{NaCl}=150.60\%=90\left(g\right)\)

f, \(m_{ddA}=\dfrac{25}{10\%}=250\left(g\right)\)

g, \(n_{NaOH}=120.20\%=24\left(g\right)\)

Gọi: n_{NaOH (thêm vào)} = a (g)

\(\Rightarrow\dfrac{a+24}{a+120}.100\%=25\%\Rightarrow a=8\left(g\right)\)

\(n_{P_2O_5}=\dfrac{99,4}{142}=0,7\left(mol\right)\)

\(P_2O_5+3H_2O\rightarrow2H_3PO_4\)

0,7        2,1           1,4 

a, \(m_{H_3PO_4}=1,4.98=137,2\left(g\right)\)

\(m_{ddH_3PO_4}=99,4+500=599,4\left(g\right)\)

Kl nước trong dd A :

\(m_{H_2O}=599,4-137,2=462,2\left(g\right)\)

\(b,C\%_{H_3PO_4}=\dfrac{137,2}{599,4}.100\%\approx22,89\%\)

\(c,C_M=\dfrac{n}{V}=\dfrac{1,4}{0,5}=2,8M\)

a) Ta có: \(C\%_{NaCl}=\dfrac{60}{60+1250}\cdot100\%\approx4,58\%\)

b) Ta có: \(C_{M_{Na_2CO_3}}=\dfrac{0,5}{0,5}=1\left(M\right)\)

\(C\%=\dfrac{30}{170}.100\%=17,647\%\) 
\(V_{\text{dd}}=\left(30+170\right)1,1=220ml\) 
\(n_{NaCl}=\dfrac{30}{58,5}=0,513mol\)
\(C_M=\dfrac{0,513}{0,22}=0,696M\)

\(C\%_{NaCl}=\dfrac{30}{170+30}.100\%=15\%\\ C_M=C\%.\dfrac{10D}{M}=10.\dfrac{10.1,1}{58,5}=1,88M\)

=50/400.100%=12,5%.

Em đăng vào môn Hoá hấy

Sửa đề: 9,2 gam Na

\(a,n_{Na_2O}=\dfrac{9,2}{23}=0,4\left(mol\right)\)

PTHH: \(Na_2O+H_2O\rightarrow2NaOH\)

            0,4------------------>0,8

\(\rightarrow C_{M\left(NaOH\right)}=\dfrac{0,8}{0,5}=1,6M\)

\(b,n_{K_2O}=\dfrac{37,6}{94}=0,4\left(mol\right)\)

PTHH: \(K_2O+H_2O\rightarrow2KOH\)

            0,4----------------->0,8

\(\rightarrow C\%_{KOH}=\dfrac{0,8.56}{362,4+37,6}.100\%=11,2\%\)