(missing in Q) : Calculate the concentration of CO & H2 & H2O when the system returns the equilibrium???
when the reaction equation is:
C(s) + H2O(g) ↔ H2(g) + CO(g)
∴ Kc = [H2] [CO] / [H2O]
and we have Kc = 0.0393 (given missing in the question)
when the O2 is added so, the reaction will be:
2H2(g) + O2(g) → 2H2O(g)
that means that 0.15 mol H2 gives 0.15 mol of H2O
∴ by using ICE table:
[H2O] [H2] [CO]
initial 0.57 + 0.15 0 0.15
change -X +X +X
Equ (0.72-X) X (0.15+X)
by substitution:
0.0393 = X (0.15+X) / (0.72-X) by solving for X
∴ X = 0.098
∴[H2] = X = 0.098 M
∴[CO] = 0.15 + X
= 0.15 + 0.098 = 0.248 M
∴[H2O] = 0.72 - X
= 0.72 - 0.098
= 0.622 M
One of the most worrisome waste products of a nuclear reactor is plutonium 239 (239Pu). This nucleus is radioactive and decays by splitting into a helium-4 nucleus and a uranium-235 nucleus (4He +... Q: One of the most worrisome waste products of a nuclear reactor is plutonium 239 (239Pu<span>).</span>
The atomic number, the number of protons and the number of electrons.
Answer:
285g of fluorine
Explanation:
To solve this problem we need to find the mass of Freon in grams. Then, with its molar mass we can find moles of freon and, as 1 mole of Freon, CCl₂F₂, contains 2 moles of fluorine, we can find moles of fluorine and its mass:
<em>Mass Freon:</em>
<em>2.00lbs * (454g / 1lb) = </em>908g of Freon
<em>Moles freon -Molar mass: 120.91g/mol- and moles of fluorine:</em>
908g of Freon * (1mol / 120.91g) =
7.5 moles of freon * (2moles Fluorine / mole Freon): 15 moles of fluorine
<em>Mass fluorine -Atomic mass: 19g/mol-:</em>
15 moles F * (19g / mol) =
<h3>285g of fluorine</h3>
