Answer:
Kc = 50.5
Explanation:
We determine the reaction:
H₂ + I₂ ⇄ 2HI
Initially we have 0.001 molesof H₂
and 0.002 moles of I₂
If we have produced 0.00187 moles of HI in the equilibrium we have to know, how many moles of I₂ and H₂, have reacted.
H₂ + I₂ ⇄ 2HI
In: 0.001 0.002 -
R: x x 2x
Eq: 0.001-x 0.002-x 0.00187
x = 0.00187/2 = 9.35×10⁻⁴ moles that have reacted
So in the equilibrium we have:
0.001 - 9.35×10⁻⁴ = 6.5×10⁻⁵ moles of H₂
0.002 - 9.35×10⁻⁴ = 1.065×10⁻³ moles of I₂
Expression for Kc is = (HI)² / (H₂) . (I₂)
0.00187 ² / 6.5×10⁻⁵ . 1.065×10⁻³ = 50.5
Answer:
<em>suface wave</em>
Explanation:
<em>hope this helps </em>
<em></em>
When trying to clean muddy, dirty river water, FILTRATION would work best
Answer is: formula of hydrate is CoCl₂· 6H₂O -c<span>obalt(II) chloride hexahydrate
</span>m(CoCl₂· xH₂O) = 1,62 g.
m(CoCl₂) = 0,88 g.
n(CoCl₂) = m(CoCl₂) ÷ M(CoCl₂)
n(CoCl₂) = 0,88 g ÷ 130 g/mol
n(CoCl₂) = 0,0068 mol.
m(H₂O) = 1,62 g - 0,88 g.
m(H₂O) = 0,74 g.
n(H₂O) = m(H₂O) ÷ m(H₂O)
n(H₂O) = 0,74 g ÷ 18 g/mol
n(H₂O) = 0,041 mol.
n(CoCl₂) : n(H₂O) = 0,0068 mol : 0,041 mol.
n(CoCl₂) : n(H₂O) = 1 : 6.
Answer:
9.47 mL
Explanation:
The reaction that takes place is:
- 2KOH + H₂SO₄ → K₂SO₄ + 2H₂O
First we <u>calculate how many KOH moles reacted</u>, using <em>the given concentration and volume of KOH solution</em>:
- 0.061 mol/L = 0.061 mmol/mL
- 0.061 mmol/mL * 26.7 mL = 1.6287 mmol KOH
Then we <u>convert KOH moles into H₂SO₄ moles</u>, using the <em>stoichiometric coefficients</em>:
- 1.6287 mmol KOH *
= 0.8144 mmol H₂SO₄
Finally we <u>calculate the required volume of the H₂SO₄ solution</u>, using<em> the number of moles and given concentration</em>:
- 0.8144 mmol ÷ 0.086 mmol/mL = 9.47 mL
