Answer:

Explanation:
Hello,
In this case, considering the reaction, we can compute the Gibbs free energy of reaction at each temperature, taking into account that the Gibbs free energy for the diatomic element is 0 kJ/mol:

Thus, at 2000 K:

And at 3000 K:

Next, since the relationship between the equilibrium constant and the Gibbs free energy of reaction is:

Thus, at each temperature we obtain:

In such a way, we can also conclude that at 2000 K reaction is unfavorable (K<1) and at 3000 K reaction is favorable (K>1).
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Answer:
please give me brainlist and follow
Explanation:
When dilute sulphuric acid is added to barium chloride solution, barium sulphate is formed as a precipitate and hydrogen chloride is formed in aqueous solution.
I believe the answer is B, as they only solely survive on tough and fibrous plant matter. Plants are producers because they can produce their own food instead of consuming something else to survive.
= k
<u>Explanation:</u>
The relation between volume, V of gas and Temperature, T of a gas is related by Charles Law.
This law states that the volume of a given amount of gas held at a constant pressure is directly proportional to the Kelvin temperature
Thus,
= k
where k is a constant
Therefore,
=
=
...
This shows, as the volume of a gas goes up, the temperature also goes up and vice-versa.
Answer:
28.0mL of the 0.0500M NaOH solution
Explanation:
<em>0.126g of lactic acid diluted to 250mL. Titrated with 0.0500M NaOH solution.</em>
<em />
The reaction of lactic acid, H₃C-CH(OH)-COOH (Molar mass: 90.08g/mol) with NaOH is:
H₃C-CH(OH)-COOH + NaOH → H₃C-CH(OH)-COO⁻ + Na⁺ + H₂O
<em>Where 1 mole of the acid reacts per mole of the base.</em>
<em />
You must know the student will reach equivalence point when moles of lactic acid = moles NaOH.
the student will titrate the 0.126g of H₃C-CH(OH)-COOH. In moles (Using molar mass) are:
0.126g ₓ (1mol / 90.08g) = <em>1.40x10⁻³ moles of H₃C-CH(OH)-COOH</em>
To reach equivalence point, the student must add 1.40x10⁻³ moles of NaOH. These moles comes from:
1.40x10⁻³ moles of NaOH ₓ (1L / 0.0500moles NaOH) = 0.0280L of the 0.0500M NaOH =
<h3>28.0mL of the 0.0500M NaOH solution</h3>