Larger elements are able to form in a supernova explosion because the star releases very large amounts of energy as well as neutrons, which allows elements heavier than iron to be produced.
<h3>What is Supernova?</h3>
This is referred to the explosion of a star and it resulting in larger elements being formed through a process known as nucleosynthesis and is usually accompanied by an increase in the brightness of the star.
The elements produced are usually larger than elements such as iron and examples include uranium, gold etc.
This is therefore the reason why it was chosen as the most appropriate choice.
Read more about Supernova here brainly.com/question/27492871
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1. Group: Groups shows the number of valence electrons and are shown in columns on the periodic table.
2. Period: Periods show the number of energy levels each element has and are shown in rows on the periodic table (up and down).
3. Blocks are show the amount of orbitals each element has. There is an s block, d block, p block, and f block.
a) After adding 10 mL of HCl
first, we need to get moles of (CH3)3N = molarity * volume
= 0.29 m * 0.025 L
= 0.00725M moles
then, we need to get moles of HCl = molarity * volume
= 0.3625 m * 0.01L
= 0.003625 moles
so moles of (CH3)3N remaining = moles of (CH3)3N - moles of HCl
= 0.00725 - 0.003625
= 0.003625 moles
and when the total volume = 0.01 L + 0.025L = 0.035 L
∴ [(CH3)3N] = moles remaining / total volume
= 0.003625 moles / 0.035L
= 0.104 M
when we have Pkb so we can get Kb :
pKb = - ㏒Kb
4.19 = -㏒Kb
∴Kb = 6.5 x 10^-5
when Kb = [(CH3)3NH+] [OH-] / [(CH3)3N]
and by using ICE table we assume we have:
[(CH3)3NH+] = X & [OH] = X
and [(CH3)3N] = 0.104 -X
by substitution:
∴ 6.5 x 10^-5 = X^2 / (0.104-X) by solving for X
∴X = 0.00257 M
∴[OH-] = X = 0.00257 M
∴POH = -㏒[OH]
= -㏒0.00257
= 2.5
∴ PH = 14 - POH
= 14 - 2.5
= 11.5
b) after adding 20ML of HCL:
moles of HCl = molarity * volume
= 0.3625 m * 0.02 L
= 0.00725 moles
the complete neutralizes of (CH3)3N we make 0.003625 moles of (CH3)3NH+ So, now we need the Ka value of (CH3)3NH+:
and when the total volume = 0.02L + 0.025 = 0.045L
∴ [ (CH3)3NH+] = moles / total volume
= 0.003625 / 0.045L
= 0.08 M
when Ka = Kw / Kb
and we have Kb = 6.5 x 10^-5 & we know that Kw = 1 x 10^-14
so, by substitution:
Ka = (1 x 10^-14) / (6.5 x 10^-5)
= 1.5 x 10^-10
when Ka expression = [(CH3)3N][H+] / [(CH3)3NH+]
by substitution:
∴ 1.5 x 10^-10 = X^2 / (0.08 - X) by solving for X
∴X = 3.5 x 10^-6 M
∴ [H+]= X = 3.5 x 10^-6 M
∴PH = -㏒[H+]
= -㏒(3.5 x 10^-6)
= 5.5
C) after adding 30ML of HCl:
moles of HCl = molarity * volume
= 0.3625m * 0.03L
= 0.011 moles
and when moles of (CH3)3N neutralized = 0.003625 moles
∴ moles of HCl remaining = moles HCl - moles (CH3)3N neutralized
= 0.011moles - 0.003625moles
= 0.007375 moles
when total volume = 0.025L + 0.03L
= 0.055L
∴[H+] = moles / total volume
= 0.007375 mol / 0.055L
= 0.134 M
∴PH = -㏒[H+]
= -㏒ 0.134
= 0.87
as water is cooled it's density increases until about 4 c then it decreases
