The correct statements which are true for the reactions in the Sun are that are fusion reactions and reactions occur because of very high temperatures.
<h3>What is nuclear fusion?</h3>
Nuclear fusion reactions are those reactions in which two species reacts with each other for the formation of new specie.
In the core of the sun, nuclear fusion of hydrogen atoms occur for the formation of helium molecules in the presence of high temperature.
Hence, the correct statements are they are fusion reactions and reactions occur because of very high temperatures.
To know more about nuclear fusion, visit the below link:
brainly.com/question/890176
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Explanation:
According to the Henderson-Hasselbalch equation,
pH = 
+ ![\frac{log[A^{-}]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7Blog%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
Given values are pH = 6, = 8
Putting given values into the above equation as follows.
6 = 8 + ![\frac{log [A^{-}]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7Blog%20%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
= -2
![\frac{[A^{-}]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E%7B-%7D%5D%7D%7B%5BHA%5D%7D)
= antilog -2
= 0.01
But according to the question, we need protonated to deprotonated ratio of ![\frac{[HA]}{[A^{-}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BHA%5D%7D%7B%5BA%5E%7B-%7D%5D%7D)
= 
= 100
Thus, we can conclude that ratio of the protonated to the deprotonated form of the acid is 
.
The answer would be D. Good luck my dude
Answer:
A reference point.
Explanation:
Reference point:
A reference point is a point which is used to determine weather the object is in motion or not. The moving object is compared with the reference point.
An object is in the state of motion when its distance is changed relative to the other object or reference point.
In order to determine weather an object is moving or not a point of reference is needed.
Without reference point displacement can not be measured
Answer:
6.574 g NaF into 300ml (0.25M HF) => Bfr with pH ~3.5
Explanation:
For buffer solution to have a pH-value of 3.5 the hydronium ion concentration [H⁺] must be 3.16 x 10⁻⁴M ( => [H⁺] = 10^-pH = 10⁻³°⁵ =3.16 x 10⁻⁴M).
Addition of NaF to 300ml of 0.25M HF gives a buffer solution. To determine mass of NaF needed use common ion analysis for HF/NaF and calculate molarity of NaF, then moles in 300ml the x formula wt => mass needed for 3.5 pH.
HF ⇄ H⁺ + F⁻; Ka = 6.6 x 10⁻⁴
Ka = [H⁺][F⁻]/[HF] = 6.6 x 10⁻⁴ = (3.16 x 10⁻⁴)[F⁻]/0.25 => [F⁻] = (6.6 x 10⁻⁴)(0.25)/(3.16x10⁻⁴) = 5.218M in F⁻ needed ( = NaF needed).
For the 300ml buffer solution, moles of NaF needed = Molarity x Volume(L)
= (5.218M)(0.300L) = 0.157 mole NaF needed x 42 g/mole = 6.574 g NaF needed.
Check using the Henderson - Hasselbalch Equation...
pH = pKa + log ([Base]/[Acid]); pKa (HF) = 3.18
Molarity of NaF = (6.572g/42g/mole)/(0.300 L soln) = 0.572M in NaF = 0.572M in F⁻.
pH = 3.18 + log ([0.572]/[0.25]) ≅ 3.5.
One can also back calculate through the Henderson -Hasselbalch Equation to determine base concentration, moles NaF then grams NaF.
