The balanced equation for the acid base reaction is as follows
NaOH + HCl ---> NaCl + H₂O
stoichiometry of NaOH to HCl is 1:1
the number of NaOH moles reacted - 0.200 mol/L x 0.0250 L = 0.005 mol
according to molar ratio
number of NaOH moles reacted = number of HCl moles reacted
therefore number of HCl moles - 0.005 mol
volume of 30.0 mL contains 0.005 mol
therefore 1000 mL contains - 0.005 mol / 0.030 L = 0.167 M
concentration of HCl is 0.167 M
According to Boyle's law, volume is inversely proportional to pressure. thus P=k/V
Therefore PV=k
P1V1=P2V2
In the question above,
P1=3.67atm
P2=1.94atm
V1=2.22L
V2=?
Thus substituting for the values in the gas equation;
3.67atm*2.22L=1.94atm*V2
V2=3.67atm*2.22L/1.94atm
=4.21L
Answer:
I think this would be iron
Explanation:
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:
Explanation:Because of the delocalised electrons exposed above and below the plane of the rest of the molecule, benzene is obviously going to be highly attractive to electrophiles - species which seek after electron rich areas in other molecules.
