Bài 6:

nP= 3,1/31= 0,1(mol)

nO2= 5/32= 0,15625(mol)

PTHH: 4P + 5 O2 -to-> 2 P2O5

Ta có: 0,1/4 < 0,15625/5

=> O2 dư, P hết , tính theo nP

=> nO2(dư)= 0,15625 - 0,1. 5/4= 1/32(mol)

=> mO2(dư)= 1/32. 32= 1(g)

b) Chất tạo thành: P2O5

nP2O5= 0,1. 2/4=0,05(mol)

=>mP2O5=142.0,05= 7,1(g)

a) -12x + 60 + 21 - 7x =15

-12x - 7x + 81 = 15

-19x = 15 - 81

-19x = -66

x= -66 : (-19)

x = \(\frac{66}{19}\)

Vậy x = \(\frac{66}{19}\)

b) 30x + 60 - 6x + 30 - 24x= 100

30x - 6x - 24x + 60 +30= 100

x ( 30 - 6- 24 ) + 90 = 100

x. 0 + 90 =100

x. 0 =10 ( vô lý )

\(\Rightarrow\) x \(\in\phi\)

k đi mình lamf cho

A = 3\(x^2\) + 9\(x\) - 7

A = 3.(\(x^2\) + 3\(x\) + \(\dfrac{9}{4}\)) - 7

A = 3.(\(x\) + \(\dfrac{3}{2}\))² - \(\dfrac{55}{4}\)

Vì (\(x\) + \(\dfrac{3}{2}\))² ≥ 0; ⇒ 3.(\(x\) + \(\dfrac{3}{2}\))² - \(\dfrac{55}{4}\) ≥ - \(\dfrac{55}{4}\)

A(min) = - \(\dfrac{55}{4}\) ⇔ \(x\) + \(\dfrac{3}{2}\) = 0 ⇔ \(x\) = - \(\dfrac{3}{2}\) 

\(\widehat{D}=80^0\)

\(\widehat{B}=135^0\)

\(4P+5O_2 \to 2P_2O_5\\ n_P=\frac{3,1}{31}=0,1(mol)\\ n_{O_2}=\frac{5}{32}=0,15625(mol)\\ \text{P hết}, O_2 \text{ dư}\\ a/ \\ m_{O_2}=(0,15625-0,125).32=1(g)\\ b/\\ \text{Chất tạo thành: } P_2O_5\\ n_{P_2O_5}=0,05(mol)\\ m_{P_2O_5}=0,05.142=7,1(g)\)

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