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emmasim [6.3K]
3 years ago
13

A closed system initially containing 1×10^-3 hydrogen 2×10^-3M iodine at 448 degree Celsius and is allowed to reach equilibrium.

The concentration of the equilibrium shows concentration of hydrogen iodude as 1.87×10^-3M.Calculate kc at this temperature
Chemistry
1 answer:
GaryK [48]3 years ago
5 0

Answer:

Kc = 50.5

Explanation:

We determine the reaction:

H₂  +  I₂   ⇄   2HI

Initially we have 0.001 molesof H₂

and 0.002 moles of I₂

If we have produced 0.00187 moles of HI in the equilibrium we have to know, how many moles of I₂ and H₂, have reacted.

           H₂     +      I₂      ⇄   2HI

In:     0.001       0.002           -

R:       x                 x                2x

Eq:  0.001-x    0.002-x      0.00187  

x = 0.00187/2 = 9.35×10⁻⁴ moles that have reacted

So in the equilibrium we have:

0.001 - 9.35×10⁻⁴ = 6.5×10⁻⁵  moles of H₂

0.002 - 9.35×10⁻⁴ = 1.065×10⁻³ moles of I₂

Expression for Kc is =  (HI)² / (H₂) . (I₂)

0.00187 ² /  6.5×10⁻⁵ . 1.065×10⁻³ = 50.5

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3 years ago
6.285×10^3 mg = _____? _____ kg
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One kilogram is equal to one thousand grams. Further, one gram is equal to 1000 mg. The conversion is as shown below,
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6. If 4 mole of the rocket fuel ammonium perchlorate, NH4C/04 (s) is
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Answer:

144g of H₂O

Explanation:

3NH₄ClO₄(s) + 3Al → Al₂O₃(s) + AlCl₃(s) + 3NO(g) + 6H₂O(g)

From the equation:

3 moles of NH₄ClO₄ produced 6 moles of H₂O

4 moles of NH₄ClO₄ produced ? moles of H₂O

(4 ₓ 6)/3 = \frac{24}{3} = 8 moles of H₂O

1 mole of H₂O = (1 × 2) + 16 = 18g (The Relative Molecular mass of H₂O)

8 moles of H₂O = ?

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Light is released when an excited electron moves from higher energy level back down to its lower, more stable energy level. The
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An insulated container contains 0.3 kg of water at 20 degrees C. An alloy with a mass of 0.090 kg and an initial temperature of
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Answer:

The specific heat of the alloy is 2.324 J/g°C

Explanation:

<u>Step 1:</u> Data given

Mass of water = 0.3 kg = 300 grams

Temperature of water = 20°C

Mass of alloy = 0.090 kg

Initial temperature of alloy = 55 °C

The final temperature = 25°C

The specific heat of water = 4.184 J/g°C

<u>Step 2:</u> Calculate the specific heat of alloy

Qlost = -Qwater

Qmetal = -Qwater

Q = m*c*ΔT

m(alloy) * c(alloy) * ΔT(alloy) = -m(water)*c(water)*ΔT(water)

⇒ mass of alloy = 90 grams

⇒ c(alloy) = the specific heat of alloy = TO BE DETERMINED

⇒ ΔT(alloy) = The change of temperature = T2 - T1 = 25-55 = -30°C

⇒ mass of water = 300 grams

⇒ c(water) = the specific heat of water = 4.184 J/g°C

⇒ ΔT(water) = The change of temperature = T2 - T1 = 25 - 20 = 5 °C

90 * c(alloy) * -30°C = -300 * 4.184 J/g°C * 5°C

c(alloy) = 2.324 J/g°C

The specific heat of the  alloy is 2.324 J/g°C

3 0
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