Write the acid-base equilibrium reaction between clo– and h2o. ignore phases.
2 answers:
Answer : The acid-base equilibrium reaction between and
will be,
Explanation :
Acid-base equilibrium reaction : It is defined as a state of chemical equilibrium that exists when the both species of conjugate acid-base pair are present in the solution.
In the reaction, an acid donates a proton to a base and to produce a conjugate acid and a conjugate base in the solution.
The acid-base equilibrium reaction between and
will be,
In this reaction, is a base and
is an acid that donates a proton to a base to produce
as a conjugate acid and
as a conjugate base.
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Answer:
7.21 × 10⁴ J
Explanation:
Ethanol is solid below -114.5°c, liquid between -114.5°C and 78.4°C, and gaseous above 78.4°C.
<em>How much heat energy is required to convert 48.3 g of solid ethanol at -114.5°C to gaseous ethanol at 135.3 °C?</em>
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We need to calculate the heat required in different stages and then add them.
The moles of ethanol are:
Solid-liquid transition
Q₁ = ΔHfus . n = (4.60 kJ/mol) . 1.05 mol = 4.83 kJ = 4.83 × 10³ J
where,
ΔHfus: molar heat of fusion
n: moles
Liquid: from -114.5°C to 78.4°C
Q₂ = c(l) . m . ΔT = (2.45 J/g.°C) . 48.3g . [78.4°C-(-114.5°C)] = 2.28 × 10⁴ J
where,
c(l): specific heat capacity of the liquid
ΔT: change in the temperature
Liquid-gas transition
Q₃ = ΔHvap . n = (38.56 kJ/mol) . 1.05 mol = 40.5 kJ = 40.5 × 10³ J
where,
ΔHvap: molar heat of vaporization
Gas: from 78.4°C to 135.3°C
Q₄ = c(g) . m . ΔT = (1.43 J/g.°C) . 48.3g . (135.3°C-78.4°C) = 3.93 × 10³ J
where
c(g): specific heat capacity of the gas
Total heat required
Q₁ + Q₂ + Q₃ + Q₄ = 4.83 × 10³ J + 2.28 × 10⁴ J + 40.5 × 10³ J + 3.93 × 10³ J = 7.21 × 10⁴ J
Answer:
Explanation:
Hello there!
In this case, since the reaction for the formation of ammonia is:
We can evidence the 1:2 mole ratio of nitrogen gas to ammonia; therefore, the appropriate stoichiometric setup for the calculation of the moles of the latter turns out to be:
And the result is:
Best regards!
H₂ (g) + F₂ (g) ⇌ 2 HF (g)
