The wrong answers for sure are B and D, I assume the answer is C
Answer:
24 atm is the total pressure exerted by the gases
Explanation:
We propose this situation:
In a vessel, we have 4 gases (for example, hydrogen, Xe, methane and chlorine)
Each of the gases has the same pressure:
6 atm → hydrogen
6 atm → xenon
6 atm → methane
6 atm → chlorine
To determine the total pressure, we sum all of them:
Partial pressure H₂ + Partial pressure Xe + Partial pressure CH₄ + Partial pressure Cl₂ = Total P
6 atm + 6 atm + 6 atm + 6 atm = 24atm
Depending on the reaction, we could monitor the progress towards equilibrium by observing the concentration of the reactant and the product are equal with time.
<h3>What is equilibrium?</h3>
Equilibrium is a stage of reaction in which the rate of forwarding reaction is equal to the rate of backward reaction and equilibrium is stable at the reversible state of mode.
The concentration of reactant and product must also be equal or the same as the time then only it can be an equilibrium reaction.
Therefore equilibrium depends on the reaction, the concentration of the reactant and the product are equal with time.
Learn more about equilibrium, here:
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Answer:
The molar concentration of Cu²⁺ in the initial solution is 6.964x10⁻⁴ M.
Explanation:
The first step to solving this problem is calculating the number of moles of Cu(NO₃)₂ added to the solution:

n = 1.375x10⁻⁵ mol
The second step is relating the number of moles to the signal. We know the the n calculated before is equivalent to a signal increase of 19.9 units (45.1-25.2):
1.375x10⁻⁵ mol _________ 19.9 units
x _________ 25.2 units
x = 1.741x10⁻⁵mol
Finally, we can calculate the Cu²⁺ concentration :
C = 1.741x10⁻⁵mol / 0.025 L
C = 6.964x10⁻⁴ M
Answer:

Explanation:
Hello,
In this case, the molar enthalpy of reaction is obtained by dividing the involved energy by the reacting moles:

Thus, it is important to notice that the compound "uses" the energy, it means that it absorbs the energy, for that reason the sign is positive. Moreover, computing the result in kJ/mol we finally obtain:

Best regards.