Answer:
How does the equilibrium change with the removal of hydrogen (H2) gas from this equation? 2H2S ⇌ 2H2(g) + S2(g) A. ... Equilibrium shifts left to produce less reactant.
Explanation:
option A is the correct answer
Equilibrium shifts right to produce more product.
I hope it will help you.
<span>When M(OH)2 dissolves we have
M(OH)2 which produces M2+ and 2OHâ’
pH + pOH=14
At ph =7; we have
7+pOH=14
pOH=14â’7 = 7
Then [OHâ’]=10^(â’pOH)
[OH-] = 10^(-7) = 1* 10^(-7)
At ph = 10. We have,
pOH = 4. And [OH-] = 10^(-4) = 1 * 10^(-4)
Finally ph = 14. We have, pOH = 0
And then [OH-] = 10^(-0) -----anything raised to zero power is 1, but (-0)...
So [OH-] = 1</span>
C unbalanced force is occuring
<span>Well it depends on percentage by what, but I'll just assume that it's percentage by mass.
For this, we look at the atomic masses of the elements present in the compound.
Cu has an atomic mass of 63.546 amu
Fe has 55.845 amu
and S has 36.065 amu
Since there are 2 molecules of Sulfur for each one of Cu and Fe, we'll multiply the Sulfur atomic weight by 2 to obtain 72.13 amu
So we have not established the mass of the compound in amus
63.546 + 55.845 + 72.13 = 191.521
That is the atomic mass of Chalcopyrite. and Iron's atomic mass is 55.845
So to get the percentage, or fraction of iron, we take 55.845 / 191.521
Which comes out to 29.15% by mass
Mass of the sample is not needed for this calculation, but since the question mentions it I would go ahead and check if the question isn't also asking for the mass of Iron in the sample as well, in which case you just find the 29.15% of 67.7g</span>
