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aalyn [17]
3 years ago
8

For+the+reaction+H2+++I2+-+2HI+the+equilibrium+constant,+kc+is+49+at+a+fixed+temperature.+Two+mole+of+hydrogen+and+two+moles+of+

iodine+are+allowed+to+reach+equilibrium+at+this+temperature.+What+is+the+concentration+of+hydrogen+iodide+at+equilibrium?
Chemistry
1 answer:
Sonja [21]3 years ago
3 0

Answer : The initial concentration of HI and concentration of HI at equilibrium is, 0.27 M and 0.386 M  respectively.

Solution :  Given,

Initial concentration of H_2 and I_2 = 0.11 M

Concentration of H_2 and I_2 at equilibrium = 0.052 M

Let the initial concentration of HI be, C

The given equilibrium reaction is,

    H_2(g)+I_2(g)\rightleftharpoons 2HI(g)

Initially               0.11   0.11            C

At equilibrium  (0.11-x) (0.11-x)   (C+2x)

As we are given that:

Concentration of H_2 and I_2 at equilibrium = 0.052 M  = (0.11-x)

The expression of K_c will be,

K_c=\frac{[HI]^2}{[H_2][I_2]}

54.3=\frac{(C+2(0.058))^2}{(0.052)\times (0.052)}

By solving the terms, we get:

C = 0.27 M

Thus, initial concentration of HI = C = 0.27 M

Thus, the concentration of HI at equilibrium = (C+2x) = 0.27 + 2(0.058) = 0.386 M

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When 47.1 J of heat is added to 14.0 g of a liquid, its temperature rises by 1.80 ∘C. What is the heat capacity of the liquid?
Alja [10]

Answer:

\boxed {\boxed {\sf 1.87 \J/g \textdegree C}}

Explanation:

We are asked to find the specific heat capacity of a liquid. We are given the heat added, the mass, and the change in temperature, so we will use the following formula.

q= mc\Delta T

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47.1 \ J = (14.0 \ g) * c * (1.80 \textdegree C)

Multiply the 2 numbers in parentheses on the right side of the equation.

47.1 \ J = (14.0 \ g * 1.80 \textdegree C)*c

47.1 \ J = (25.2 \ g*\textdegree C) *c

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\frac {47.1 \ J}{(25.2 g *\textdegree C)} = \frac {(25.2 g *\textdegree C)*c}{{(25.2 g *\textdegree C)}}

\frac {47.1 \ J}{(25.2 g *\textdegree C)} =c

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The original measurements of heat, mass, and temperature all have 3 significant figures, so our answer must have the same. For the number we found that is the hundredth place. The 9 in the thousandth place to the right tells us to round the 6 up to a 7.

1.87 \ J/ g * \textdegree C =c

The heat capacity of the liquid is approximately 1.87 J/g°C.

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