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ololo11 [35]
3 years ago
9

How much heat is required to decompose 25.5 grams of NaHCO3? 2NaHCO3(s) + 129 kJ 2Na2CO3(s) + H2O(

Chemistry
1 answer:
Vaselesa [24]3 years ago
4 0
To determine the heat required in order to decompose a certain amount of a substance, we need information on the heat needed to decompose one mole of the substance. This value are readily available online and other sources. For this reaction, the heat needed is 129 kJ per 2 mol of NaHCO3. We calculate as follows:

129 KJ / 2 mol NaHCO3 (1 mol / 84.01 g ) (25.5 g NaHCO3 ) = 19.58 kJ of heat is needed.
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ki77a [65]

Answer:

1.15 atm

Explanation:

According to Dalton's law of partial pressures, the total pressure is the sum of all the partial pressures of the gases present in the mixture.

Therefore we have:

Total pressure = partial pressure of carbon monoxide + partial pressure of oxygen + partial pressure of carbon dioxide

We were given the following:

Total pressure = 2.45 atm

Pressure of oxygen = 0.65 atm

Pressure of carbon monoxide = x

Pressure of carbon dioxide = 0.65 atm

Therefore:

2.45 = x + 0.65 + 0.65

2.45 = x + 1.3

x = 2.45 - 1.3

x = 1.15 atm

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Imagine mixing 1 tablespoon of Epsom salt with 2 cups of ammonia. How much precipitate would be produced? Describe the amount of
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Answer:

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6 0
3 years ago
Given the following information, what is the concentration of H2O(g) at equilibrium? [H2S](eq) = 0.671 M [O2](eq) = 0.587 M Kc =
MAVERICK [17]

<u>Answer:</u> The equilibrium concentration of water is 0.597 M

<u>Explanation:</u>

Equilibrium constant in terms of concentration 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 a general chemical reaction:

aA+bB\rightleftharpoons cC+dD

The expression for K_{eq} is written as:

K_{c}=\frac{[C]^c[D]^d}{[A]^a[B]^b}

The concentration of pure solids and pure liquids are taken as 1 in the expression.

For the given chemical reaction:

2H_2S(g)+O_2(g)\rightleftharpoons 2S(s)+2H_2O(g)

The expression of K_c for above equation is:

K_c=\frac{[H_2O]^2}{[H_2S]^2\times [O_2]}

We are given:

[H_2S]_{eq}=0.671M

[O_2]_{eq}=0.587M

K_c=1.35

Putting values in above expression, we get:

1.35=\frac{[H_2O]^2}{(0.671)^2\times 0.587}

[H_2O]=\sqrt{(1.35\times 0.671\times 0.671\times 0.587)}=0.597M

Hence, the equilibrium concentration of water is 0.597 M

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