Answer:
[H₂] = 1.61x10⁻³ M
Explanation:
2H₂S(g) ⇋ 2H₂(g) + S₂(g)
Kc = 9.30x10⁻⁸ =
First we <u>calculate the initial concentration</u>:
0.45 molH₂S / 3.0L = 0.15 M
The concentrations at equilibrium would be:
[H₂S] = 0.15 - 2x
[H₂] = 2x
[S₂] = x
We <u>put the data in the Kc expression and solve for x</u>:
We make a simplification because x<<< 0.0225:
x = 8.058x10⁻⁴
[H₂] = 2*x = 1.61x10⁻³ M
Answer:
14.533 grams of solid precipitate of mercury(II) dichromate will form.
Explanation:
Moles of mercury(II) acetate =
Moles of sodium dichromate =
According to reaction , 1 mole of sodium dichromate reacts with 1 mole of mercury(II) acetate , then 0.045906 moles of sodium dichromate will recat with :
of mercury(II) acetate
This means that mercury(II) acetate is present in an excess amount and sodium dichromate is present in limiting amount.So, amount of precipitate will depend upon moles of sodium dichromate.
According to reaction , 1 mole of sodium dichromate gives 1 mole of mercury(II) dichromate , then 0.045906 moles of sodium dichromate will give :
of mercury(II) dichromate
Mass of 0.045906 moles of mercury(II) dichromate:
0.045906 mol × 316.59 g/mol = 14.533 g
14.533 grams of solid precipitate of mercury(II) dichromate will form.
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Explanation: Look at the scrrn shot and you can see what is on there or not. :)))))
