Answer:
K remains the same;
Q < K;
The reaction must run in the forward direction to reestablish the equilibrium;
The concentration of will decrease.
Explanation:
In this problem, we're adding an excess of a reactant, chlorine gas, to a system that is already at equilibrium. According to the principle of Le Chatelier, when a system at equilibrium is disturbed, the equilibrium shifts toward the side of the equilibrium that minimizes the disturbance.
Since we'll have an excess of chlorine, the system will try to reduce that excess by shifting the equilibrium to the right. Therefore, the reaction must run in the forward direction to reestablish the equilibrium.
The value of K remains the same, as it's only temperature-dependent, while the value of Q will be lower than K, that is, Q < K, as Q < K is the case when reaction proceeds to the right.
As a result, since is also a reactant, its concentration will decrease.
Answer:
<em>C. The electron-withdrawing fluorine atoms pull electron density from the oxygen in trifluoroacetate. The negative charge is more stabilized in trifluoroacetate by this effect.</em>
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Explanation:
<em>The structures of trifluoroacetate and acetic acid are both shown in the image attached.</em>
<em>The trifluoroacetate anion (CF3CO2-), just like the acetate anion has in the middle, two oxygen atoms.</em>
<em>However, in the trifluoroacetate anion, there are also three electronegative fluorine atoms attached to the nearby carbon atom attached to the carbonyl, and these pull some electron density through the sigma bonding network away from the oxygen atoms, thereby spreading out the negative charge further. This effect, called the "inductive effect" stabilizes the anion formed,the trifouoroacetate anion is thus more stabilized than the acetate anion.</em>
<em>Hence, trifluoroacetic acid is a stronger acid than acetic acid, having a pKa of -0.18.</em>
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<u><em>Hope this helps!</em></u>
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Percent (%) Composition of CuO
Cu = 1 x 50g - Multiply by one as there is one Cu
O = 1 x 12.5g - Multiply by one as there is one O
CuO = 62.5g
% for Cu = 50g over 62.5 multiplied by 100 = 80%
% for O = 12.5g over 62.5 multiplied by 100 = 20%
Final Answer :
<em>Percent (%) Composition of CuO = </em>80% (Cu) & 20% (O)
Answer:
Explanation:
To convert atoms to moles, Avogadro's Number must be used: 6.022*10²³.
This tells us the amount of particles (atoms, molecules, etc.) in 1 mole of a substance. In this case it is the atoms of potassium. We can create a ratio.
Multiply by the given number of moles: 15.2
The moles of potassium cancel.
The denominator of 1 can be ignored.
Multiply.
The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated that is the hundredth place. The 3 in the thousandth place tells us to leave 5.
In 15.2 moles of potassium, there are <u>9.15*10²⁴ atoms of potassium.</u>
Answer:
A:temperature
Explanation:
The temperature cannot be determined by looking at the spectra of the star due to lack of the equipment for its measurement. <em>On the other-hand, the remaining statements like the distance from earth, movement towards or away from earth can be determined.</em>