M(NO3) = 14+3*16 = 62
W(N) = 14/62 = 0.23 =23%
W(O) = 48/62 = 0.77 = 77%
M(SO3) = 32+3*16 = 80
W(S) = 32/80 = 0.4 = 40%
W(O) = 48/80 = 0.6 = 60%
M(H2SO4) = 2*1+32+4*16 = 98
W(H) = 2/98 = 0.02 = 2%
W(S) = 32/98 = 0.33 = 33%
W(O) = 64/98 = 0.65 = 65%
4Mn + 10HNO3 = 4Mn(NO3)2 + NH4NO3 + 3H2O
Mn(0) -2e- --> Mn(+2) I 4
N(+5) +8e- ----> N(-3) I 1
Hg+2H2SO4=HgSO4+SO2+2H2O
Hg(0)-2e---->Hg(+2)
S(+6)+2e---->S(+4)
Х г 39 г
2Al + Cr2O3 = Al2O3 + 2Cr
n = 2 моль n = 1 моль
Mr = 27 г/моль Mr = 152 г/моль
m = 54 г m = 152 г
Х г Al - 39 г Cr2O3
54 г Al - 152 г Cr2O3
m(Al) = 54 * 39 / 152 = 13,8 г
CH₃-CH₂-CH₂-OH - пропанол-1
CH₃-CH-CH₃ - пропанол-2
|
OH
CH₃-CH₂-CH₂-CH₂-OH - бутанол-1
CH₃-CH-CH₂-CH₃ - бутанол-2
|
OH
<span><span>Дано:
m(</span>Ca(OH)</span>₂<span>) = 7,4 мг = 0,0074 г
Найти: n(</span>Ca(OH)₂<span>)
Решение:
M(</span>Ca(OH)₂<span>) = 40 + 16</span>·2<span> +
1</span>·2<span>= 74 г\моль
n(</span>Ca(OH)₂<span>) = m\M = 0,0074г\74 г\моль= 0,0001</span><span><span> моль
Ответ: 0</span>,0001 моль</span>