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Ksenya-84 [330]
4 years ago
15

An atom with an atomic number of 18 will have how many electrons in its innermost orbit

Chemistry
1 answer:
Paraphin [41]4 years ago
6 0

Answer:

2 electrons

Explanation:

The K shell is the innermost orbit and it contains 2 electrons

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What is the formal charge of nitrogen in no2cl
Gre4nikov [31]

Answer:

The formal charge on nitrogen in NO_{2}Cl is +1.

Explanation:

The structure of NO_{2}Cl is as follows.

(In attachment)

Formal\, charge = Valence\, electron - (nonbonding\, electrons + \frac{Bonding\,electrons}{2})

From the structure, Nitrogen has no non bonding electrons. Nitrogen has four bonds and each bond corresponds to 2 electrons. Hence, nitrogen have eight bonding electrons and five valence electrons.

Formal\,charge\,on\,nitrogen = 5-[0+ \frac{8}{2}]= +1

Therefore, The formal charge on nitrogen in NO_{2}Cl is +1.

6 0
3 years ago
The molecule carbon disulfide (CS2) is nonpolar and has only London dispersion forces between the molecules. Carbon tetrachlorid
stepladder [879]

Answer:

Final Answer: boiling point and high

Explanation:

3 0
4 years ago
Read 2 more answers
Draw the resonance structures for the conjugate base of Phenol (C6H6O). In one sentence, explain why phenol (C6H6O) has a pKa of
notsponge [240]

Answer:

The four resonance structures of the phenoxide ion are shown in the image attached

The conjugate base of cyclohexanol has only one resonance contributor, while

the conjugate base of phenol has four resonance contributors.

Explanation:

In organic chemistry, it is known that structures are more stable if they possess more resonance contributors. The greater the number of contributing canonical structures, the more stable the organic specie. Since the phenoxide ion has four contributing canonical structures, it is quite much more stable than cyclohexanol having only one contributing structure to its conjugate base. Hence the PKa(acid dissociation constant) of phenol is lesser than that of cyclohexanol. The conjugate base of phenol is stabilized by resonance.

6 0
3 years ago
At a given temperature, the elementary reaction A − ⇀ ↽ − B , A↽−−⇀B, in the forward direction, is first order in A A with a rat
Svetllana [295]

Answer:

The equilibrium constant for the reversible reaction = 0.0164

Explanation:

At equilibrium the rate of forward reaction is equal to the rate of backwards reaction.

The reaction is given as

A ⇌ B

Rate of forward reaction is first order in [A] and the rate of backward reaction is also first order in [B]

The rate of forward reaction = |r₁| = k₁ [A]

The rate of backward reaction = |r₂| = k₂ [B]

(Taking only the magnitudes)

where k₁ and k₂ are the forward and backward rate constants respectively.

k₁ = 0.010 s⁻¹

k₂ = 0.0610 s⁻¹

|r₁| = 0.010 [A]

|r₂| = 0.016 [B]

At equilibrium, the rate of forward and backward reactions are equal

|r₁| = |r₂|

k₁ [A] = k₂ [B] (eqn 1)

Note that equilibrium constant, K, is given as

K = [B]/[A]

So, from eqn 1

k₁ [A] = k₂ [B]

[B]/[A] = (k₁/k₂) = (0.01/0.0610) = 0.0163934426 = 0.0164

K = [B]/[A] = (k₁/k₂) = 0.0164

Hope this Helps!!!

5 0
3 years ago
The atomic masses of 151eu and 153eu are 150.919860 and 152.921243 amu, respectively. The average atomic mass of europium is 151
vitfil [10]

Answer:-  The natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for ^1^5^1_E_u and ^1^5^3_E_u as 150.919860 and 152.921243 amu, respectively.  Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of ^1^5^1_E_u as n then the abundance of ^1^5^3_E_u would be 1-n .

Let's plug in the values in the formula:

151.964=150.919860(n)+152.921243(1-n)

151.964=150.919860n+152.921243-152.921243n

on keeping similar terms on same side:

151.964-152.921243=150.919860n-152.921243n

-0.957243=-2.001383n

negative sign is on both sides so it is canceled:

0.957243=2.001383n

n=\frac{0.957243}{2.001383}

n=0.478

The abundance of ^1^5^1_E_u is 0.478 which is 47.8%.  

The abundance of ^1^5^3_E_u is = 1-0.478

= 0.522 which is 52.2%

Hence, the natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .


3 0
3 years ago
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