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.
Answer:
8.08 × 10⁻⁴
Explanation:
Let's consider the following reaction.
COCl₂(g) ⇄ CO (g) + Cl₂(g)
The initial concentration of phosgene is:
M = 2.00 mol / 1.00 L = 2.00 M
We can find the final concentrations using an ICE chart.
COCl₂(g) ⇄ CO (g) + Cl₂(g)
I 2.00 0 0
C -x +x +x
E 2.00 -x x x
The equilibrium concentration of Cl₂, x, is 0.0398 mol / 1.00 L = 0.0398 M.
The concentrations at equilibrium are:
[COCl₂] = 2.00 -x = 1.96 M
[CO] = [Cl₂] = 0.0398 M
The equilibrium constant (Keq) is:
Keq = [CO].[Cl₂]/[COCl₂]
Keq = (0.0398)²/1.96
Keq = 8.08 × 10⁻⁴
Answer:
Uh first of all this is algebra but I'll answer this
First distribute the three and 5 (Multiply them by both terms inside parenthesis.
3x-6=5x+20
Then add like terms
8x=14
Divide 8 by 8 and 8 by 14
x = 14/8
Explanation:
diatomic hydrogen is written as H2 (2.02 grams H2) <------- if each hydrogen atom is 1.01 grams, then two hydrogen atoms are 2.02 grams 2.0 moles H2 X 2.02 grams H2 ------------- (divide to cancel moles) = 4.04 grams/mole H2 ÷ one mole = 4.04 grams H2
Answer:
combustion is a high-temperature exothermic redox chemical reaction between a fuel and an oxidant, usually atmospheric oxygen, that produces oxidized, often gaseous products, in a mixture termed as smoke.
