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jek_recluse [69]
3 years ago
9

When the following reaction comes to equilibrium, will the concentrations of the reactants or products be greater? Does the answ

er to this question depend on the initial concentrations of the reactants and products? A(g)+B(g)⇌2C(g)Kc=1.4×10−5 Match the words in the left column to the appropriate blanks in the sentences on the right. ResetHelp The value of the equilibrium constant is relatively ; therefore, the concentration of reactants will be the concentration of products. This result is the initial concentration of the reactants and products.
Chemistry
1 answer:
Nutka1998 [239]3 years ago
4 0

Answer:

At equilibrium, the concentration of the reactants will be greater than the concentration of the products. This does not depend on the initial concentrations of the reactants and products.

Explanation:

The value of Kc gives us an idea of the extent of the reaction. A big Kc (Kc > 1) means that in the equilibrium there are more products than reactants, and the opposite happens for a small Kc (Kc < 1). The equilibrium is reached no matter what the initial concentrations are.

The value of the equilibrium constant is relatively SMALL; therefore, the concentration of reactants will be GREATER THAN the concentration of products. This result is INDEPENDENT OF the initial concentration of the reactants and products.

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the balanced chemical equation for the reaction is : -

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3 0
3 years ago
Air trapped in a cylinder fitted with a piston occupies 142.8 mL at 0.97 kPa pressure.
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Answer:

92.344mL

Explanation:

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5 0
3 years ago
A flask with a volume of 125.0 mL contains air with a density of 1.298 g/L. what is the mass of the air contained in the flask?
horsena [70]
A flask with a volume of 125.0 mL contains air with a density of 1.298 g/L. what is the mass of the air contained in the flask<span>The given are: </span>
<span><span>1.      </span>Mass = ?</span><span><span /></span>
<span><span>2.      </span>Density = 1298 g/L</span>
3.      Volume = 125mL to L
a. 125 ml x 0.001l/1ml = 0.125 L

<span>Formula and derivation: </span><span><span>
1.      </span>density = mass / volume</span> <span><span>
2.      mass </span>= density / volume</span>

<span>Solution for the problem: </span><span><span>

1. mass = </span></span> <span> 1298 g/L / 0.125 L = 10384g
</span>


8 0
3 years ago
The following equation is one way to prepare oxygen in a lab. 2KClO3 → 2KCl + 3O2 Molar mass Info: MM O2 = 32 g/mol MM KCl = 74
Alex73 [517]
From  the  equation  above   the  reacting   ratio  of  KClO3   to  O2  is  2:3 therefore  the  number  of  moles  of  oxygen  produced  is  ( 4 x3)/2 =  6 moles  since   four  moles  of  KClO3  was  consumed
mass=relative  formula mass  x  number  of  moles
That  is   32g/mol x 6  moles  =192grams


8 0
3 years ago
Read 2 more answers
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