№1
SO₂ M(S)=32 M(0)=16 M(SO₂)=32+16*2=64
a)W(S)=32/64*100%=50%
b)M(SO3)=32+16*3=80
80>64
c)W(S)=32/80*100%=40% W(O)=16*3/80=48/80=60%
№2
a)SixOy W(Si)=46,7%=0,467
M(Si)=28 M(O)=16 M(SixOy)=28/0,467=60
60/28=2,1
60/16=3,75(округлим)
Si:O=2:4=1:2
SiO₂
b)M(SiO2)=28+16*2=60
№3
a)2Mg+O₂=2MgO
b)6Na+N₂=2Na₃N
c)2Al+3S=Al₂S₃
№4
a)В результате реакции молекулы метана с двумя молекулами кислорода образовалось две молекулы воды и одна молекула углекислого газа.
b)CH₄+2O₂=CO₂+2H₂O
№5
а)3MgO+2H₃PO₄=Mg₃(PO₄)₂+3H₂O Реакция обмена
b)2Al+3H₂SO₄=Al₂(SO₄)₃+3H₂ Реакция замещения
There is a formula to calculate pH of buffer made of weak acid (here it's NaH2PO4) and its salt with strong base (Na2HPO4):
pH = pKa+lg CM Na2HPO4/CM NaH2PO4;
let's take that 450 ml of buffer is sum of volumes of Na2HPO4 and NaH2PO4 solutions (V1 ml - volume of Na2HPO4 sol. and V2 ml - volume of NaH2PO4 sol.), i.e. V1+V2 = 450 ml = 0.45 l. (1);
as CM = n/V and CM both of salts = 1 mol/l, so we have following: CM Na2HPO4 = V1/0.45 = 2.222*V1 and CM NaH2PO4 = V2/0.45 = 2.222*V2;
lg 2.222*V1/2.222*V2 = pH-pKa = 6.4-6.8 = -0.4, so 2.222*V1/2.222*V2 = 10^-0.8 = 0.3981;
as V2 = 0.45-V1 (see (1) above), we get 2.222*V1/2.222*(0.45-V1) = 0.3981, so V1 = 0.128 l. or 128 ml;
V2 = 0.45-0.128 = 0.322 l. or 322 ml;
Volume of Na2HPO4 sol. is 128 ml.;
Volume of NaH2PO4 sol. is 322 ml.
