Answer:
The rate of the forward reaction and the rate of the reverse reaction are equal
The concentrations of the reactants and products no longer change.
Explanation:
The equilibrium state is a state of rest or motion of chemical system.
These are some of the condition for dynamic equilibrium of a system:
the systems must be involved in reversible chemical reactions.
the rate of forward reaction is equal to that of the reverse process.
the system offers no resistance to any change in any of the factors.
there is no change in concentration of each of the species in equilibrium with respect to time.
We can see that a system in equilibrium clearly identifies with the chosen options.
Answer: 61 grams
Explanation:
To calculate the number of moles, we use the equation:
The chemical equation for the combustion of octane in oxygen follows the equation:
By stoichiometry of the reaction;
25 moles of oxygen react with 2 moles of octane
4.69 moles of oxygen react with=
moles of octane
Thus, oxygen is the limiting reagent as it limits the formation of product and octane is the excess reagent.
25 moles of oxygen produce 18 moles of water
4.69 moles of oxygen produce=
moles of water.
Mass of water produced=
The maximum mass of water that could be produced by the chemical reaction is 61 grams.
Robert Hooke was the scientist that coined the term cell
Answer:
Hydroxide concentration of the sample is 1.3x10⁻⁶M
Explanation:
The equilibrium constant of water, Kw, is:
H₂O(l) ⇄ H⁺(aq) + OH⁻(aq)
Kw is defined as:
Kw = 1.7x10⁻¹² = [H⁺] [OH⁻]
As the sample is of pure water, both H⁺ and OH⁻ ions have the same concentration because come from the same equilibrium, that is:
[H⁺] = [OH⁻]
We can write the Kw expression:
1.7x10⁻¹² = [OH⁻] [OH⁻]
1.7x10⁻¹² = [OH⁻]²
1.3x10⁻⁶M = [OH⁻]
<h3>Hydroxide concentration of the sample is 1.3x10⁻⁶M</h3>
The combustion reaction of octane is as follow,
C₈H₁₈ + 25/2 O₂ → 8 CO₂ + 9 H₂O
According to balance equation,
8 moles of CO₂ are released when = 114.23 g (1 mole) Octane is reacted
So,
6.20 moles of CO₂ will release when = X g of Octane is reacted
Solving for X,
X = (114.23 g × 6.20 mol) ÷ 8 mol
X = 88.52 g of Octane
Result:
88.52 g of Octane is needed to release 6.20 mol CO₂.
