The last one, answer is D
Answer:
Dark matter makes up 85% of the mass of the universe. Dark matter is not directly observable because it doesn't interact with any electromagnetic wave. In the development of the universe, without dark matter, the universe will not function, move or rotate as it does now (this speculation led to the quest to find the anomaly of mass and energy in the known universe, eventually leading to the idealization of dark matter) and will not have enough gravitational force to hold it together. After the big bang,<em> the presence of dark matter and energy ensured that the newly formed universe didn't just float away, rather, it provided enough gravitational force to hold the universe while still allowing it to expand sufficiently</em>.
The development of the universe would have been different without the universe in the sense that the young universe won't have enough mass to hold it together, and the universe would have simply floated apart. The behavior of the universe would have been different from what we observe now, and some physical laws that applies now will not apply to the universe.
Answer:
Hydrofluoric acid.
Explanation:
To know which of the acid is the strongest, let us determine the pka of each acid. This is illustrated below:
1. Acetic acid
Ka = 1.8x10^-5
pKa =..?
pKa = –logKa
pKa = –Log 1.8x10^-5
pKa = 4.74
2. Benzoic acid
Ka = 6.5x10^-5
pKa =..?
pKa = –logKa
pKa = –Log 6.5x10^-5
pKa = 4.18
3. Hydrofluoric acid.
Ka = 6.8x10^-4
pKa =..?
pKa = –logKa
pKa = –Log 6.8x10^-4
pKa = 3.17
4. Hypochlorous acid
Ka = 3.0x10^-8
pKa =..?
pKa = –logKa
pKa = –Log 3.0x10^-8
pKa = 7.52
Note: the smaller the pKa value, the stronger the acid.
The pka of the various acids as calculated above is given below:
Acid >>>>>>>>>>>>>>>>>> pKa
1. Acetic acid >>>>>>>>>> 4.74
2. Benzoic acid >>>>>>>> 4.18
3. Hydrofluoric acid >>>> 3.17
4. Hypochlorous acid >> 7.52
From the above illustration, we can see that hydrofluoric acid has the lowest pKa value. Therefore, hydrofluoric acid is the strongest among them.
Answer:
Change in internal energy (ΔU) = -9 KJ
Explanation:
Given:
q = –8 kJ [Heat removed]
w = –1 kJ [Work done]
Find:
Change in internal energy (ΔU)
Computation:
Change in internal energy (ΔU) = q + w
Change in internal energy (ΔU) = -8 KJ + (-1 KJ)
Change in internal energy (ΔU) = -8 KJ - 1 KJ
Change in internal energy (ΔU) = -9 KJ
The swimming pools pH is below 7, meaning it is slightly acidic. If you want to make the pH higher, you must add a base which by definition has a pH higher than 7.
D. Add base
