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2 years ago

With what elements does nitrogen form ionic bonds?

Chemistry

1 answer:

Andrew [12]2 years ago

5 0

Answer:

Group 1 and 2 elements

Explanation:

Nitrogen, a non-metal will form ionic bonds with most group 1 and group 2 metals on the periodic table.

How does ionic bonds form?

They are bonds formed between a highly electronegative specie and one with very low electronegativity.
As such, ionic bonds forms between metals and non-metals
In this bond type, the metal due to its electropositive nature will transfer electrons to the non-metals for it to gain.
The non-metals becomes negatively charged as the metal is positively charged.
The electrostatic attraction between the two specie leads to the formation of ionic bonds.

Most metals in group 1 and 2 fits in this description. Some of them are calcium, magnesium, lithium, Barium e.t.c.

It mostly favors group 2 metals.

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A transition in the balmer series for hydrogen has an observed wavelength of 434 nm. Use the Rydberg equation below to find the

DanielleElmas [232]

Answer:

i. n = 5

ii. ΔE = 7.61 × $10^{-46}$ KJ/mole

Explanation:

1. ΔE = (1/λ) = -2.178 × $10^{-18}$ ( $\frac{1}{n^{2}_{final} }$ - $\frac{1}{n^{2}_{initial} }$ )

(1/434 × $10^{-9}$ ) = -2.178 × $10^{-18}$ ( $\frac{n^{2}_{initial} - n^{2}_{final} }{n^{2}_{final} n^{2}_{initial} }$ )

⇒ 434 × $10^{-9}$ = (1/-2.178 × $10^{-18}$ ) $\frac{n^{2}_{final} *n^{2}_{initial} }{n^{2}_{initial} - n^{2}_{final} }$

But, $n_{final}$ = 2

434 × $10^{-9}$ = (1/2.178 × $10^{-18}$ ) $\frac{2^{2} n^{2}_{initial} }{n^{2}_{initial} - 2^{2} }$

434 × $10^{-9}$ × 2.178 × $10^{-18}$ = $(\frac{4n^{2}_{initial} }{n^{2}_{initial} - 4 })$

⇒ $n_{initial}$ = 5

Therefore, the initial energy level where transition occurred is from 5.

2. ΔE = hf

= (hc) ÷ λ

= (6.626 × 10−34 × 3.0 × $10^{8}$ ) ÷ (434 × $10^{-9}$ )

= (1.9878 × $10^{-25}$ ) ÷ (434 × $10^{-9}$ )

= 4.58 × $10^{-19}$ J

= 4.58 × $10^{-22}$ KJ

But 1 mole = 6.02× $10^{23}$ , then;

energy in KJ/mole = (4.58 × $10^{-22}$ KJ) ÷ (6.02× $10^{23}$ )

= 7.61 × $10^{-46}$ KJ/mole

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Answer: 3

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Equation for the reaction of decay of $_{19}^{40}\textrm{K}$ radioisotope follows:

$Moles of zirconium=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{15.4}{91}=0.17moles$

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$_{41}^{91}\textrm{Nb}\rightarrow _{40}^{91}\textrm{Zr}+_{+1}^0e$

By the stoichiometry of above reaction,

1 mole of $_{40}^{91}\textrm{Zr}$ is produced by 1 mole $_{41}^{91}\textrm{Nb}$

So, 0.17 moles of $_{40}^{91}\textrm{Zr}$ will be produced by = $\frac{1}{1}\times 0.17=0.17\text{ moles of }_{40}^{91}\textrm{Nb}$

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decomposed will be = 0.17 moles

Initial amount of $_{40}^{91}\textrm{Nb}$ will be = Amount decomposed + Amount left = (0.17 + 0.024)moles =0.194 moles

$a=\frac{a_o}{2^n}$

where,

a = amount of reactant left after n-half lives = 0.024

$a_o$ = Initial amount of the reactant = 0.194

n = number of half lives= ?

Putting values in above equation, we get:

$0.024=\frac{0.194}{2^n}$

$n=3$

Therefore, 3 half lives have passed.

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The product of a reaction between these two elements is $Al_{2} O_{3}$ .

Explanation:

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The oxide is having charge of -2

The product of these reactants will produce a chemical compound.

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