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kompoz [17]
4 years ago
5

An atom of argon in the ground state tends not to bond with an atom of a different element because the argon atom has (a)a total

of two valence electrons (b)more protons than neutrons (c)more neutrons than protons or (d)a total of eight valence electrons
Chemistry
2 answers:
bezimeni [28]4 years ago
7 0

Answer: Option (d) is the correct answer.

Explanation:

Atomic number of argon is 18 and its electronic configuration is 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}.

Also, it is known that number of electrons in K, L and M shells of argon are 2, 8, 8.

Thus, we can conclude that an atom of argon in the ground state tends not to bond with an atom of a different element because the argon atom has a total of eight valence electrons.

So, as per the octet rule, last sub-shell of argon is completely filled. Hence, it will not form any bond with any other atom.

Dvinal [7]4 years ago
5 0
The answer is <span>a total of eight valence electrons
hope this help
</span>
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How many miles are there in 27 grams of aluminum?
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7 0
3 years ago
Imagine that you have a 7.00 L gas tank and a 3.50 L gas tank. You need to fill one tank with oxygen and the other with acetylen
GaryK [48]

Answer:

We should fill the acetylene tank to a pressure of 100 atm

Explanation:

Step 1: Data given

Volume tank 1 = 7.00 L

Tank 1 is filled with oxygen and has a pressure of 125 atm

Volume tank 2 = 3.50 L

Step 2: The balanced equation

2 C2H2 + 5 O2 → 4 CO2 + 2 H2O

Step 3: Calculate the pressure of acetylene

p(oxygen) * V(Oxygen) = n(Oxygen) * R * T  

p(Acetylene) * V(Acetylene) = n(Acetylene) * R * T

We can assume both tanks are at the same temperature, so we can write this as followed:

p(Oxygen)* V(Oxygen) / p(Acetylene)*V(Acetylene) = n(Oxygen) / n(Acetylene)    

⇒with n(Oxygen) / n(Acetylene) = 5/2

⇒with p(Oxygen) = 125atm

⇒V(Oxygen) =7L

⇒V(Acetylene) = 3.5L

⇒ this gives us:   (125 * 7) / (P(Acetylene) * 3.5) = 5/2

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We should fill the acetylene tank to a pressure of 100 atm

7 0
3 years ago
If for a particular process, ΔH=308 kJmol and ΔS=439 Jmol K, in what temperature range will the process be spontaneous?
Tamiku [17]

Answer:

The process will be spontaneous above 702 K.

Explanation:

Step 1: Given data

  • Standard enthalpy of the reaction (ΔH°): 308 kJ/mol
  • Standard entropy of the reaction (ΔS°): 439 J/mol.K

Step 2: Calculate the temperature range in which the process will be spontaneous

The reaction will be spontaneous when the standard Gibbs free energy (ΔG°) is negative. We can calculate ΔG° using the following expression.

ΔG° = ΔH° - T × ΔS°

When ΔG° < 0,

ΔH° - T × ΔS° < 0

ΔH° < T × ΔS°

T > ΔH°/ΔS°

T > (308,000 J/mol)/(439 J/mol.K)

T > 702 K

The process will be spontaneous above 702 K.

7 0
3 years ago
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