The reaction is (a reação é)
(C6H4(OH)2(aq) + H2O2(aq) --> C6H4O2(aq) + H2O(L) ........(i)
The other reaction given are (A outra reação dada é)
C6H4(OH)2(aq) ---> C6H4(OH)2(aq) + H2(g) .......(ii) ΔH = 177 kJ
2H2O2(aq) ----> 2H2O(l) + O2(g) ..................................(iii) ΔH = -189.3
2H2(g) + O2(g) ----> 2H2O (l) .....................................(iv) ΔH =-572 kJ
We can obtain equation (i) from equation (ii), (iii) and (iv) as
[Podemos obter a equação (i) da equação (ii), (iii) e (iv) como]
equation (i) = equation (ii) + 0.5 X equation (iii) + 0.5 X equation (iv)
[equação (i) = equação (ii) + 0.5 X equação (iii) + 0.5 X equação (iv)]
So ΔH1 = ΔH(ii) + 0.5 X ΔH(iii) + 0.5 X ΔH(iv)
= 177 + 0.5 X (-189.3) + 0.5 X(-572) = -203.65 kJ
Answer:
maybe try 4.002602, good luck
Answer: 500Pa
Explanation:
Using Boyle's law which is a gas law, it states that the pressure and volume of a gas have an inverse relationship. That is to say, If volume increases, then pressure decreases and vice versa, when temperature is held constant. Therefore, when the volume is halved, the pressure is doubled; and if the volume is doubled, the pressure is halved.
So in this case, since the temperature is held at constant (off) the value of the pressure is half the value of the volume plunger, being 500Pa
Answer:
molecular structure of pectin with labeling
Explanation:
