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pochemuha
3 years ago
6

Which of the following pairs of atoms do NOT exhibit an ionic bond?

Chemistry
2 answers:
Eddi Din [679]3 years ago
7 0
N and H are both gases and would not form an ionic bond.
Olegator [25]3 years ago
4 0

Answer:

F and Cu.

Explanation:

Hello,

In this case, by knowing that the farther the elements in the periodic table, the more ionic they are, in addition to the difference in the electronegativity which must be greater than 1.7 for the bond to be ionic, we can substantiate that the iron and the copper don't exhibit an ionic bond as shown below:

\Delta E=E_{Cu}-E_{Fe}=1.9-1.83=0.07

Such value is characteristic for covalent bonds.

Best regards.

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How do I find the slope of this graph?​ Is it even possible?
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Not exactly But you can take the slope of the curved portion and the slope of the flatline.

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4 0
2 years ago
Potassium hydrogen phthalate is a solid, monoprotic acid frequently used in the laboratory as a primary standard. It has the unw
Tomtit [17]

Answer:

0.1082M of Barium Hydroxide

Explanation:

KHP reacts with Ba(OH)2 as follows:

2KHP + Ba(OH)2 → 2H2O + Ba²⁺ + 2K⁺ + 2P²⁻

<em>Where 2 moles of KHP reacts per mole of barium hydroxide</em>

<em />

To solve this question we must find the moles of KHP in 1.37g. With these moles and the reaction we can find the moles of Ba(OH)2 and its molarity using the volume of the solution (31.0mL = 0.0310L) as follows:

<em>Moles KHP -Molar mass: 204.22g/mol-</em>

1.37g * (1mol / 204.22g) = 0.006708 moles KHP

<em>Moles Ba(OH)2:</em>

0.006708 moles KHP * (1mol Ba(OH)2 / 2mol KHP) =

0.003354 moles Ba(OH)2

<em>Molarity:</em>

0.003354 moles Ba(OH)2 / 0.0310L =

<h3>0.1082M of Barium Hydroxide</h3>
4 0
3 years ago
You are performing a titration of a triprotic acid, when you spill water on your lab notebook. you can read that: pka 1 = 1.40,
eimsori [14]
According to the PH formula:
PH= Pka +㏒ [strong base/weak acid]
when we have PH at the first equivalence =3.35 and the Pka1 = 1.4
So, by substitution, we can get the value of ㏒[strong base / weak acid]
3.35 = 1.4 + ㏒[strong base/ weak acid]
∴㏒[strong base/weak acid] = 3.35-1.4 = 1.95 
to get the Pka2 we will substitute with the value of ㏒[strong base/ weak acid] and the value of PH of the second equivalence point
∴Pk2 = PH2 - ㏒[strong base/ weak acid]
          = 7.55 - 1.95 = 5.6 
5 0
3 years ago
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