Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Trong $Ca_3(PO_4)_3 : $
$\%P = \dfrac{31.3}{310}.100\% = 30\%$
Trong $Ca(H_2PO_4)_2: $
$\%P = \dfrac{31.2}{234}.100\% = 26,5\%$
Trong $(NH_4)_3PO_4 : $
$\%P = \dfrac{31}{149}.100\% = 20,8\%$
Vậy $\%P$ : $Ca_3(PO_4)_2 > Ca(H_2PO_4)_2 > (NH_4)_3PO_4$
em cảm ơn ạ nhưng mà anh chép sai đầu bài của em mất rồi:). dù gì em vẫn cảm ơn ạ.
*Hợp chất A
Theo gt:%XA=75%
=>\(\dfrac{X}{X+4Y}\).100%=75%
=>100X=75X+300Y
=>25X=300Y=>X=12Y
*Hợp chất B
Gọi CTTQ B là:XaYb
a:b=\(\dfrac{90\%}{X}:\dfrac{10\%}{Y}\)=\(\dfrac{90\%}{12Y}:\dfrac{10\%}{Y}\)=3:4
=>CTHH B là:X3Y4
Bài 1 :
a) Gọi $n_{Mg} = a ; n_{Ca} = b \Rightarrow 24a + 40b = 20,8(1)$
Ta có :
$\%m_{Mg} = \dfrac{24a + 24a}{20,8 + 24a}.100\% = 37,5\%(2)$
Từ (1)(2) suy ra a = 0,2 ; b = 0,4
b)
Bảo toàn Ca : $n_{Ca_3(PO_4)_2} = \dfrac{1}{3}n_{Ca} = \dfrac{0,4}{3}(mol)$
$\Rightarrow m_{Ca_3(PO_4)_2} = \dfrac{0,4}{3}.310 = 41,33(gam)$
a) \(n_{Fe_3O_4}=\dfrac{46,4}{232}=0,2\left(mol\right)\)
=> mFe = 0,2.3.56 = 33,6 (g)
=> mO = 0,2.4.16 = 12,8 (g)
b) \(n_{Ca_3\left(PO_4\right)_2}=\dfrac{12,4}{310}=0,04\left(mol\right)\)
=> mCa = 0,04.3.40 = 4,8 (g)
=> mP = 0,04.2.31 = 2,48(g)
=> mO = 0,04.8.16 = 5,12(g)
PTHH: \(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\\ 0,1mol:0,2mol\rightarrow0,2mol:0,1mol:0,1mol\)
\(n_{Na_2CO_3}=\dfrac{10,6}{106}=0,1\left(mol\right)\)
a. \(n_{NaCl}=0,2.58,5=11,7\left(g\right)\)
b. \(V_{CO_2}=0,1.22,4=2,24\left(l\right)\)
c. \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)
\(n_{Na_2CO_3}=\dfrac{m_{Na_2CO_3}}{M_{Na_2CO_3}}=\dfrac{10,6}{106}=0,1\left(mol\right)\)
PTHH:
\(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\)
\(0,1............0,2........0,2.......0,1.......0,1\) (mol)
a. \(m_{NaCl}=M_{NaCl}.n_{NaCl}=0,2.58,5=11,7\left(g\right)\)
b. \(V_{CO_2}=22,4.n_{CO_2}=22,4.0,1=2,24\left(l\right)\)
c. \(m_{HCl}=n_{HCl}.M_{HCl}=0,2.36,5=7,3\left(g\right)\)
*CuO
\(\%M_{Cu}=\frac{64}{80}.100\%=80\%\)
\(\%M_O=100-80=20\%\)
*P2O5
\(\%M_P=\frac{31.2}{142}.100\%=43,66\%\)
\(\%M_O=100-43,66=56,34\%\)
*\(H2SO4\)
\(\%M_H=\frac{2}{98}.100\%=2,04\%\)
\(\%M_S=\frac{32}{98}.100\%=32,65\%\)
\(\%M_O=100-32,65-2,04=65,31\%\)
* Al2(SO4)3
\(\%M_{Al}=\frac{27.2}{400}.100\%=13,5\%\)
\(\%M_S=\frac{32.3}{400}.100\%=24\%\)
\(\%M_O=100-24-13,5=62,5\%\)
\(\cdot NH4NO3\)
\(\%M_N=\frac{14.2}{80},100\%=35\%\)
\(\%M_H=\frac{4}{80}.100\%=5\%\)
\(\%M_O=100-35-5=60\%\)
*\(Ca3\left(PO4\right)3\)
\(\%M_{Ca}=\frac{40.3}{405},100\%=29,85\%\)
\(\%M_P=\frac{31.3}{405}.100\%=22,96\%\)
\(\%M_O=100-22,96-29,85=47,19\%\)
CuO: 80%C, 20%O.
P2O5: 43,66%P; 47,34%O
H2SO4: 2%H; 32,7%S; 65,3%O
Al2(SO4)3: 15,8%Al; 28,1%S; 56,1%O
NH4NO3: 35%N; 5%H; 60%O
Ca3(PO4)2: 38,85Ca; 20%P; 41,15%O
\(a.\)
\(n_{Na_2CO_3}=\dfrac{10.6}{106}=0.1\left(mol\right)\)
\(n_{Na}=0.1\cdot2=0.2\left(mol\right)\)
\(\Rightarrow\%Na=\dfrac{0.2\cdot23}{10.6}\cdot100\%=43.39\%\)
\(n_C=0.2\left(mol\right)\)
\(\Rightarrow\%C=\dfrac{0.1\cdot12}{10.6}\cdot100\%=11.32\%\)
\(n_O=0.2\cdot3=0.6\left(mol\right)\)
\(\Rightarrow\%O=\dfrac{0.6\cdot16}{10.6}\cdot100\%=45.29\%\)
\(b.\)
\(n_{Ca_3\left(PO_4\right)_2}=\dfrac{620}{310}=2\left(mol\right)\)
\(n_{Ca}=3\cdot2=6\left(mol\right)\)
\(\Rightarrow\%Ca=\dfrac{6\cdot40}{620}\cdot100\%=38.71\%\)
\(n_P=2\cdot2=4\left(mol\right)\)
\(\Rightarrow\%P=\dfrac{4\cdot31}{620}\cdot100\%=20\%\)
\(n_O=2\cdot8=16\left(mol\right)\)
\(\Rightarrow\%O=\dfrac{16\cdot16}{620}\cdot100\%=41.29\%\)