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3 years ago

Which of the following is not a diatomic molecule? nitrogen hydrogen bromine helium

Chemistry

2 answers:

Nikitich [7]3 years ago

7 0

Answer: Option (d) is the correct answer.

Explanation:

When two atoms are bonded together then it is known as a diatomic molecule.

Such as nitrogen exists as $N_{2}$ molecule in atmosphere so, it is a diatomic molecule.

Hydrogen also exists as $H_{2}$ molecule in atmosphere so, it is also a diatomic molecule. Also, bromine is a diatomic molecule because it exists as $Br_{2}$ molecule into the atmosphere.

But helium exists as He molecule as it has 2 electrons so, as per the octet rule it's s-shell is completely filled. Hence, it is stable and exists as a monoatomic molecule into the atmosphere.

Thus, we can conclude that out of the given options, helium is not a diatomic molecule.

vovangra [49]3 years ago

7 0

Answer: Helium

Explanation:

Diatomic molecule is one which is formed by the combination of two atoms.

Nitrogen exists as $N_2$ in atmosphere, Hydrogen exists as $H_2$ in atmosphere, Bromine exists as $Br_2$ in atmosphere and Helium exists as He in atmosphere.

Helium being a inert gas or noble gas do not combine with another atom as its octet is already complete and thus it is unreactive.

He with atomic number 10 has following configuartion:- $1s^22s^22p^6$

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If 24.3 g of NO and 13.8 g of O₂ are used to form NO₂, how many moles of excess reactant will be left over?2 NO (g) + O₂ (g) → 2

zhuklara [117]

Explanation:

2 NO (g) + O₂ (g) ----> 2 NO₂ (g)

24.3 g of NO are reacting with 13.8 g of O₂. First we can convert the mass of theses samples into moles using their molar masses.

molar mass of O = 16.00 g/mol

molar mass of N = 14.01 g/mol

molar mass of NO = 16.00 g/mol + 14.01 g/mol

molar mass of NO = 30.01 g/mol

molar mass of O₂ = 2 * 16.00 g/mol

molar mass of O₂ = 32.00 g/mol

moles of NO = 24.3 g * 1 mol/(30.01 g)

moles of NO = 0.810 moles

moles of O₂ = 13.8 g * 1 mol/(32.00 g)

moles of O₂ = 0.431 moles

Now, to determine the limiting reactant or the excess reactant we can find the number of moles of O₂ that will react with 0.810 moles of NO and the number of moles of NO that will react with 0.431 moles of O₂.

According to the coefficients of the reaction 2 moles of NO will react with 1 mol of O₂. Let's use that relationship to find the limiting reagent.

2 moles of NO = 1 mol of O₂

moles of O₂ = 0.810 moles of NO * 1 mol of O₂/(2 moles of NO)

moles of O₂ = 0.405 moles

moles of NO = 0.431 moles of O₂ * 2 moles of NO/(1 mol of O₂)

moles of NO = 0.862 moles

We found that we need 0.405 moles of O₂ to completely react with 0.810 moles of NO. Or, we need 0.862 moles of NO to completely react with ours 0.431 moles of NO.

We can say that NO is limiting our reaction and O₂ is in excess.

Only 0.405 moles of O₂ will react with 0.810 moles of NO. But we had 0.431 moles of it. Let's find the excess.

Excess of O₂ = 0.431 moles - 0.405 moles

Excess of O₂ = 0.026 moles

Answer: 0.026 moles is the number of moles of oxygen that will be left over.

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