Answer: That would be false because it is the contact between two layers representing a gap in the geologic record, usually from the erosion of the layers which would normally be expected to appear.
Explanation:
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Answer:
i believe that the answer is c
Explanation:
I think its c because because its used as a bed rock layer
Types of Bonds can be predicted by calculating the
difference in electronegativity.
If, Electronegativity difference is,
Less
than 0.4 then it is Non Polar Pure Covalent
Between 0.4 and 1.7 then it is Polar Covalent
Greater than 1.7 then it is Ionic
For Br and Br,
E.N of Bromine = 2.96
E.N of Bromine = 2.96
________
E.N Difference
0.00 (Non Polar/Pure Covalent)
For N and O,
E.N of Oxygen = 3.44
E.N of Nitrogen = 3.04
________
E.N Difference
0.40 (Non Polar/Pure Covalent)
For P and H,
E.N of Hydrogen = 2.20
E.N of Phosphorous = 2.19
________
E.N Difference 0.01 (Non Polar/Pure Covalent)
For K and O,
E.N of Oxygen = 3.44
E.N of Potassium = 0.82
________
E.N Difference 2.62 (Ionic)
You missed a lot of details in your question, so when we have the complete question as the attached picture so, the answer would be:
when we have the value of Ka = 1.8 x 10^-5 so, we can use it to get the Pka value
by using this formula:
Pka = -㏒Ka
= -㏒(1.8 x 10^-5)
= 4.7
now, after we have got the Pka we need now to get moles of NaC2H3O2 and
moles of HC2H3O2:
when moles of NaOH = 0.015 moles
when moles NaC2H3O2 after adding NaOH
= initial mol NaC2H3O2 + mol NaOH
∴moles NaC2H3O2 = 0.1 + 0.015 = 0.115 moles
and moles HC2H3O2 after adding NaOH
= initial mol HC2H3O2 - mol NaOH
∴ moles HC2H3O2 = 0.1 - 0.015 = 0.085 moles
so, when we have moles [HC2H3O2] &[NaC2H3O2] so we can substitution its values in [A] &[HA] :
by using H-H equation we can get the PH:
when PH = Pka + ㏒[A]/[HA] PH = 4.7 + ㏒0.115/0.085
= 4.8