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2 years ago

Consider a possible triatomic molecule containing a nitrogen, oxygen, and fluorine atoms for the following questions.

Chemistry

1 answer:

STALIN [3.7K]2 years ago

5 0

Answer:

a) nitrogen

b) nitrogen =5

Oxygen = 6

Fluorine =7

Explanation:

Usually, if we have two or more elements in a compound, the central atom in the compound is the atom having the least value of electro negativity.

If we consider fluorine, oxygen and nitrogen; nitrogen is the least electronegative of the trio hence it should be the central atom of the triatomic molecule.

The number of valence electrons on the valence shell of each atom is shown below;

nitrogen =5

Oxygen = 6

Fluorine =7

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Hydrogen

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Hydrogen can never be central atom despite its low electronegativity

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A box of jello has a mass of 250 g. How many boxes must be bought to have 1 Kg of jello? *

zhuklara [117]

Answer:

Number of boxes = 4

Explanation:

Given:

Mass of one box of jello = 250 grams

Total quantity want to purchase = 1 kg = 1 × 1,000 gram = 1,000 grams

Find:

Number of boxes in 1,000 grams = ?

Computation:

Number of boxes = Total quantity want to purchase / Mass of one box of jello

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2 years ago

Consider the insoluble compound nickel(II) hydroxide , Ni(OH)2 . The nickel ion also forms a complex with cyanide ions . Write a

natali 33 [55]

Answer: Equilibrium constant for this reaction is $2.8 \times 10^{15}$ .

Explanation:

Chemical reaction equation for the formation of nickel cyanide complex is as follows.

$Ni(OH)_{2}(s) + 4CN^{-}(aq) \rightleftharpoons [Ni(CN)_{4}^{2-}](aq) + 2OH^{-}(aq)$

We know that,

K = $K_{f} \times K_{sp}$

We are given that, $K_{f} = 1.0 \times 10^{31}$

and, $K_{sp} = 2.8 \times 10^{-16}$

Hence, we will calculate the value of K as follows.

K = $K_{f} \times K_{sp}$

K = $(1.0 \times 10^{31}) \times (2.8 \times 10^{-16})$

= $2.8 \times 10^{15}$

Thus, we can conclude that equilibrium constant for this reaction is $2.8 \times 10^{15}$ .

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2 years ago

Use the periodic table to write the electron configuration for rubidium (Rb) in noble has notation

Alexxx [7]

Answer:

Rb: [Kr] 5s

Step-by-step explanation:

Rb is element 37, the first element in Period 5.

It has one valence electron, so its valence electron configuration is 5s.

The noble gas configuration uses the symbol of the previous noble gas as a shortcut for the electron configurations of the inner electrons.

The preceding noble gas is Kr, so the electron configuration is Rb: [Kr] 5s.

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3 years ago