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Ksenya-84 [330]
3 years ago
9

Why is the oxidation state of o2- more likely than o6+ please explain ?

Chemistry
1 answer:
Alex Ar [27]3 years ago
3 0

<u>Answer:</u> Because more energy is required to form O^{6+} ion.

<u>Explanation:</u>

Oxygen is the 8th element of the periodic table which belongs to Group 16.

The electronic configuration of the element is: 1s^22s^22p^4

The number of valence electrons are 6 which is not a stable electronic configuration.

To attain stable electronic configuration, this element will either loose 6 electrons to form O^{6+} ion or gain 2 electrons to form O^{2-} ion.

The loosing of 6 electrons require a huge amount of energy and is an unfavorable reaction. Hence, this element will gain 2 electrons easily and form O^{2-} ion.

This element is considered as a non-metal because it gains electron to attain stable electronic configuration.

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Asparagine is a polar amino acid shown here at ph 7 what is the maximum
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Maximum number of water molcules which can take part in forming hydrogen bonding interaction with asparagine are 13. Following structure shows the interactions. Lone pair of electrons act as hydrogen bond acceptors and hydrogen atoms attached to heteroatoms acts as hydrogen bond donors.

3 0
3 years ago
The titration of Na2CO3 with HCl has the following qualitative profile: a. Identify the major species in solution at points A-F.
exis [7]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Note: This question is incomplete and lacks very important data to solve this question. But I have found the similar question which shows the profiles about which question discusses. Using the data from that question, I have solved the question.

a) We need to find the major species from A to F.

Major Species at A:

1. Na_{2} CO_{3}

Major Species at B:

1. Na_{2} CO_{3}

2. NaHCO_{3}

Major Species at C:

1. NaHCO_{3}

Major Species at D:

1. NaHCO_{3}

2. H_{2}CO_{3}

Major Species at E:

1. H_{2}CO_{3}

Major Species at F:

1. H_{2}CO_{3}

b) pH calculation:

At Halfway point B:

pH = pKa_{1} + log[CO_{3}.^{-2}]/[HCO_{3}.^{-1}]

pH = pKa_{1} = 6.35

Similarly, at halfway point D.  

At point D,

pH = pKa_{2} + log [HCO_{3}.^{-1}]/[H2CO_{3}]

pH = pKa_{2} = 10.33

8 0
3 years ago
The density of lead is 11.4g/cm3. What volume, in ft3, would be occupied by 10.0 g of lead?
yawa3891 [41]

Answer:

3.10×10¯⁵ ft³.

Explanation:

The following data were obtained from the question:

Density (D) of lead = 11.4 g/cm³

Mass (m) of lead = 10 g

Volume (V) of lead =.?

Density (D) = mass (m) / Volume (V)

D = m/V

11.4 = 10 / V

Cross multiply

11.4 × V = 10

Divide both side by 11.4

V = 10 / 11.4

V = 0.877 cm³

Finally, we shall convert 0.877 cm³ to ft³. This can be obtained as follow:

1 cm³ = 3.531×10¯⁵ ft³

Therefore,

0.877 cm³ = 0.877 cm³ × 3.531×10¯⁵ ft³ /1 cm³

0.877 cm³ = 3.10×10¯⁵ ft³

Thus, 0.877 cm³ is equivalent to 3.10×10¯⁵ ft³.

Therefore, the volume of the lead in ft³ is 3.10×10¯⁵ ft³.

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(4.9 x 10-2) (9.80 x 102) =
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Gas is the answer ok
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