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3 years ago

What is the net amount of energy released when one mole of h2o(?) is produced?

1 answer:

german3 years ago

7 0

The net amount of energy produced can be obtained from a table of enthalpy change of formation, available online.

The enthalpy change of formation indicate how much energy the 1 mole of the product (H2O) has relative to the elemental reactants (H2 and O2). In other words, the "lost" energy equals the heat/energy released.

For water (H2O), this value is -285.8 if the final product is a liquid under standard conditions, and -241.82 if the product is in gas form which contains some energy that could be further released. This means that if the final product (H2O) is in liquid form, energy released is 285.8 kJ/mol.

Since water is in liquid form under standard conditions, the first value (285.8 kJ/mol) is generally appropriate.

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A radioactive material, with half-life of six months, has 100 thousand unstable nuclei.

goldfiish [28.3K]

Answer:

See Explanation

Explanation:

Given that;

N/No = (1/2)^t/t1/2

Where;

No = amount of radioactive isotope originally present

N = A mount of radioactive isotope present at time t

t = time taken

t1/2 = half life

N/1000=(1/2)^3/6

N/1000=(1/2)^0.5

N = (1/2)^0.5 * 1000

N= 707 unstable nuclei

Since the value of the initial activity of the radioactive material was not given, the activity of the radioactive material after three months is given by;

Decay constant = 0.693/t1/2 = 0.693/6 months = 0.1155 month^-1

Hence;

A=Aoe^-kt

Where;

A = Activity after a time t

Ao = initial activity

k = decay constant

t = time taken

A = Aoe^-3 *0.1155

A=Aoe^-0.3465

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3 years ago

Read 2 more answers

A certain reaction with an activation energy of 185 kJ/mol was run at 505 K and again at 525 K . What is the ratio of f at the h

frosja888 [35]

Answer:

The ratio of f at the higher temperature to f at the lower temperature is 5.356

Explanation:

Given;

activation energy, Ea = 185 kJ/mol = 185,000 J/mol

final temperature, T₂ = 525 K

initial temperature, T₁ = 505 k

Apply Arrhenius equation;

$Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]$

Where;

$\frac{f_2}{f_1}$ is the ratio of f at the higher temperature to f at the lower temperature

R is gas constant = 8.314 J/mole.K

$Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]\\\\Log(\frac{f_2}{f_1} ) = \frac{185,000}{2.303 \times 8.314} [\frac{1}{505} -\frac{1}{525} ]\\\\Log(\frac{f_2}{f_1} ) = 0.7289\\\\\frac{f_2}{f_1} = 10^{0.7289}\\\\\frac{f_2}{f_1} = 5.356$