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4 years ago

Hydrogen and chlorine react to form hydrogen chloride, like this:

Chemistry

1 answer:

Lyrx [107]4 years ago

6 0

Answer: The value of the equilibrium constant for this reaction is 2.7

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_c$

For the given chemical reaction:

$H_2(g)+Cl_2(g)\rightarrow 2HCl(g)$

The expression for $K_c$ is written as:

$K_c=\frac{[HCl]^2}{[H_2]^1[I_2]^1}$

$K_c=\frac{(1.6)^2}{(0.78)^1\times (1.2)^1}$

$K_c=2.7$

Thus the value of the equilibrium constant for this reaction is 2.7

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What is larger than a galaxy?

Ray Of Light [21]

the milky way is much larger than a galaxy

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3 years ago

A 10 gram sample of water is heated to 105 ℃ and is mixed with a 25 gram sample of water cooled to 25℃ . What is the final tempe

lesya692 [45]

Answer:

The final temperature of the water mixture is 47.85°C

Explanation :

Given,

For Warm Water

mass = 10grams

Temperature = 105°C

For Cold Water

mass = 25grams

Temperature = 25°C

When a sample of warm water is mixed with a sample of cool water,

The energy amount going out of the warm water is equal to the energy amount going into the cool water. This means:

<h3>Qlost = Qgain</h3>

However,

Q = (mass) (ΔT) (Cp)

Cp = Specific heat of water = 4.184 J/Kg°C

So,

(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)

We start by calling the final, ending temperature 'x.' Keep in mind that BOTH water samples will wind up at the temperature we are calling 'x.' Also, make sure you understand that the 'x' we are using is FINAL temperature. This is what we are solving for.

The warmer water goes down from to 105°C to x, so this means its Δt equals 105°C − x. The colder water goes up in temperature, so its Δt equals x − 25℃

Substituting the values,

(10)( 105°C − x)(4.184) = (25)(x − 25℃)(4.184)

Solving for x, we get

x = 47.85°C

Therefore, The final temperature of the water mixture is 47.85°C.

8 0

4 years ago

Suppose that H2SO4 was used in the reaction instead of HCl. How many moles of NaOH would neutralize 1 mole of H2SO4?

arlik [135]

Answer: Hence, 2 moles of NaOH would neutralize 1 mole of $H_2SO_4$

Explanation:

$HCl+NaOH\rightarrow NaCl+H_2O$

One mole of hydrochloric acid neutralizes one mole of sodium hydroxide to give one mole of sodium chloride and one mole of water

When sulfuric acid is used in the place of HCl , 1 mole of sulfuric acid will neutralizes the 2 mole of sodium hydroxide and gives one mole of sodium sulfate and 2 moles of water. As we can see from the reaction:

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$