Repeatable
If it’s repeatable and you get the same answer constantly then it’s a good and accurate experiment
Answer:
27.64 liters
Explanation:
From the balanced equation, 2 moles of K2Cr2O7 requires 3 moles of CH3OH.
Mole of CH3OH = 1.9/32.04 = 0.0593 mole
Mole of K2Cr2O7 that will require 0.0593 mole of CH3OH:
2 x 0.0593/3 = 0.0395 mole
mole = molarity x volume
Volume of K2Cr2O7 needed = 0.0395/0.00143
= 27.64 Liter
<em>Hence, 27.64 liters of 0.00143 M K2Cr2O7 will be required to titrate 1.90 g of CH3OH dissolved in 50.0 mL of solution</em>
Answer:
Answers are in the explanation.
Explanation:
<em>Given concentrations are:</em>
- <em>SO₂ = 0.20M O₂ = 0.60M SO₃ = 0.60M</em>
- <em>SO₂ = 0.14M O₂ = 0.10M SO₃ = 0.40M </em>
- <em>And SO₂ = 0.90M O₂ = 0.50M SO₃ = 0.10M</em>
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In the reaction:
2SO₂(g) + O₂(g) ⇄ 2SO₃(g)
Kc is defined as:
Kc = 15 = [SO₃]² / [O₂] [SO₂]²
<em>Where concentrations of each species are equilbrium concentrations.</em>
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Also, you can define Q (Reaction quotient) as:
Q = [SO₃]² / [O₂] [SO₂]²
<em>Where concentrations of each species are ACTUAL concentrations.</em>
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If Q > Kc, the reaction will shift to the left until Q = Kc;
If Q < Kc, the reaction will shift to the right until Q = Kc
If Q = Kc, there is no net reaction because reaction would be en equilibrium.
Replacing with given concentrations:
- Q = [0.60M]² / [0.60M] [0.20M]² = 15; Q = Kc → No net reaction
- Q = [0.40M]² / [0.10M] [0.14M]² = 82; Q > Kc, → Reaction will shift to the left
- Q = [0.10M]² / [0.50M] [0.90M]² = 0.015; Q < Kc → Reaction will shift to the right
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Answer:
aphelion
Explanation:
A planet moves with constantly changing speed as it moves about its orbit. The fastest a planet moves is at perihelion (closest) and the slowest is at aphelion (farthest).
