<u>Answer:</u> It repels positive ion and attracts negative ion.
<u>Explanation:</u>
There are 2 types of ions:
1. Cations: These ions are positively charged ions which are formed when a substance looses electrons.
2. Anions: These ions are negatively charged ions which are formed when a substance gains electrons.
It is known that, like charges repel each other and unlike charges attract each other.
As, it is given that the substance is positively charged, so it will attract anion and repel cation.
Answer:
8 mi = 12.875 km
12.875 /60 m
= 0.21458 km / min
11.7 / 0.21458 = 54.52 min
Its d. Speed (v) and frequency(f) both stay the same.
<span>Because Amplitude does not affect wave frequency. When you put more energy into the wave (larger amplitude) it does not affect how often you repeat a cycle.
</span>Also speed can be altered when the medium changes through which it travels.
Answer: The volume of 0.235 M 
needed to titrate 40.0 mL of 0.0500 M 
is 8.51 ml
Explanation:
To calculate the volume of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is
are the n-factor, molarity and volume of base which is
We are given:
Putting values in above equation, we get:
Thus volume of 0.235 M needed to titrate 40.0 mL of 0.0500 M is 8.51 ml
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
