Whiteish silver
shiny
ductile
conductor
metal
left
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
Answer:
When light is shone on to the surface of a metal, its electrons absorb small amounts of energy and become excited into one of its many empty orbitals. The electrons immediately fall back down to lower energy levels and emit light. This process is responsible for the high luster of metals.
Explanation:
<em> </em><em>Your </em><em>well-wisher</em><em> </em><em>:-)</em>
The balanced reaction is given as:
<span>2Ca(s)+O2(g)→2CaO(s)
</span>
We are given the amount of oxygen gas to be used for the reaction. This will be the starting point of our calculations.
0.016 mol O2 ( 2 mol CaO / 1 mol O2 ) = 0.032 mol CaO
<span>Therefore, for the reaction, the amount of calcium oxide produced is 0.032 moles.</span>
