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Natalija [7]
3 years ago
12

According to their locations on the Periodic Table, argon (Ar) and neon (Ne) are

Chemistry
2 answers:
beks73 [17]3 years ago
4 0

Answer: Option (D) is the correct answer.

Explanation:

Atomic number of neon is 10 and its electronic distribution is 2, 8. On the other hand, atomic number of argon is 18 and its electronic distribution is 2, 8, 8.

As both these elements have completely filled orbital and they have 8 valence electrons. Hence, both of them belongs to group 18 which is also known as noble gas group.

In the Periodic table, neon lies in period 2 and group 18. On the other hand, argon lies in period 3 and group 18.

Thus, we can conclude that according to their locations on the Periodic Table, argon (Ar) and neon (Ne) are noble gases.

a_sh-v [17]3 years ago
3 0
Your answer would be (D), Noble Gases, or Inert Gases - The far right on the periodic table is also known as Group(0)., or Group 18 on the periodic table. Elements in this group includes, Neon(NE), Argon(AR), Krypton(KR), Xenon(XE), and Radon(RN).





Hope that helps!!!
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Answer:the answer is C

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7 0
3 years ago
Which of the following leads to a higher rate of diffusion?
tiny-mole [99]
<h3><u>Answer;</u></h3>

Higher velocity of particles

<h3><u>Explanation;</u></h3>

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5 0
3 years ago
The decomposition of HBr(g) into elemental species is found to have a rate constant of 4.2 ×10−3atm s−1. If 2.00 atm of HBr are
Dennis_Churaev [7]

Answer:

7,94 minutes

Explanation:

If the descomposition of HBr(gr) into elemental species have a rate constant, then this reaction belongs to a zero-order reaction kinetics, where the r<em>eaction rate does not depend on the concentration of the reactants. </em>

For the zero-order reactions, concentration-time equation can be written as follows:

                                          [A] = - Kt + [Ao]

where:

  • [A]: concentration of the reactant A at the <em>t </em>time,
  • [A]o: initial concentration of the reactant A,
  • K: rate constant,
  • t: elapsed time of the reaction

<u>To solve the problem, we just replace our data in the concentration-time equation, and we clear the value of t.</u>

Data:

K = 4.2 ×10−3atm/s,  

[A]o=[HBr]o= 2 atm,  

[A]=[HBr]=0 atm (all HBr(g) is gone)

<em>We clear the incognita :</em>

[A] = - Kt + [Ao]............. Kt =  [Ao] - [A]

                                        t  = ([Ao] - [A])/K

<em>We replace the numerical values:</em>

t = (2 atm - 0 atm)/4.2 ×10−3atm/s = 476,19 s = 7,94 minutes

So, we need 7,94 minutes to achieve complete conversion into elements ([HBr]=0).

6 0
3 years ago
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Bas_tet [7]

Answer:

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Explanation:

I hope this helps!!!!!!

3 0
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