A 2.200-g sample of quinone (C6H4O2) is burned in a bomb calorimeter whose total heat capacity is 7.854<span> kJ/°C. The temperature of the calorimeter increases from 23.44 to </span>30.57 °C<span>. </span>
Answer:
At equilibrium:
[H2] = 0.005 M
[Br2] = 0.105 M
[HBr] = 0.189 M
Explanation:
H2(g) + Br2(g) ⇄ 2HBr
an "x" value will be used from reactant to produced "2x"
so at equilibrium:
[H2] = 0.1 - x
[Br2] = 0.2 - x
[HBr] = 2x
we know that Kc=[HBr]²/[H2][Br2]
Thus 62.5 = (2x)²/(0.1-x)(0.2-x)
this generate a quadratic equation: 58.5x² - 18.75x + 1.25 = 0
the x₁ = 0.23 x₂ = 0.09457
we pick 0.09457 because the two reactants can not make more than what they have. x₁ is higher than both initial reactant concentration
Then we substitute the "x₂" value at equilibrium:
[H2] = 0.1-0.09457 = 0.005 M
[Br2] = 0.2-0.09457 = 0.105 M
[HBr] = 2*0.09457 = 0.189 M
The answer is natural gasses
D. mixture in which its components retain their identity
In a heterogeneous mixture, multiple substances are mixed, but they don't chemically react and they remain chemically the same.