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k0ka [10]
3 years ago
15

Consider the equation below. Upper C a upper C upper O subscript 3 (s) double-headed arrow upper C a upper O (s) plus upper C up

per O subscript 2 (g). What is the equilibrium constant expression for the given reaction? K subscript e q equals StartFraction bracket upper C a upper O EndBracket over upper C a upper C upper O subscript 3 EndBracket EndFraction. K subscript e q equals StartFraction StartBracket upper C upper O subscript 2 EndBracket StartBracket upper C a upper O EndBracket over StartBracket upper C a upper C upper O subscript 3 EndBracket EndFraction. K subscript e q equals StartBracket upper C upper O subscript 2 EndBracket. K subscript e q equals StartFraction 1 over StartBracket upper C upper O subscript 2 EndBracket EndFraction.
Chemistry
2 answers:
stiks02 [169]3 years ago
8 0

Answer:K subscript e q equals StartFraction StartBracket upper C upper O subscript 2 EndBracket StartBracket upper C a upper O EndBracket over StartBracket upper C a upper C upper O subscript 3 EndBracket EndFraction

Explanation: the answer has it's root in Law of mass action which states that; the rate of a chemical reaction is directly proportional to the product of the concentrations of the reactants raised to their respective stoichiometric coefficients.

Len [333]3 years ago
8 0

Answer: is A

Explanation:pij

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13.50 grams of Pb(NO3)4 are dissolved in enough water to make 250 mL of solution. What is the molar its of the resulting solutio
likoan [24]
Data:
M (molarity) = ? (M or Mol/L)
m (mass) = 13.50 g
V (volume) = 250 mL → 0.25 L
MM (Molar Mass) of Lead(IV) Nitrate Pb(NO_3)_4
Pb = 1*207 = 207 amu
N = (1*14)*4 = 14*4 = 56 amu
O = (3*16)*4 = 48*4 = 192 amu
------------------------------------
MM of Pb(NO_3)_4 = 207+56+192 = 455 g/mol

Formula:
M =  \frac{m}{MM*V}

Solving:
M = \frac{m}{MM*V}
M =  \frac{13.50}{455*0.25}
M =  \frac{13.50}{113.75}
M = 0.118681318...\:\:\to\:\:\boxed{\boxed{M \approx 0.119\:Mol/L}}\end{array}}\qquad\quad\checkmark

Answer:
<span>B. 0.119 M</span>
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<h3>What is pH?</h3>

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