Covalent bonds are strong bonds. Atoms that share pairs of electrons form molecules. A molecule is a group of atoms held together by covalent bonds. A diatomic molecule is a molecule containing only two atoms.
Answer:
85.34g of NH3
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
N2 + 3H2 —> 2NH3
Step 2:
Determination of the number of moles of NH3 produced by the reaction of 2.51 moles of N2. This is illustrated below:
From the balanced equation above,
1 mole of N2 reacted to produce 2 moles of NH3.
Therefore, 2.51 moles of N2 will react to produce = (2.51 x 2)/1 = 5.02 moles of NH3.
Therefore, 5.02 moles of NH3 is produced from the reaction.
Step 3:
Conversion of 5.02 moles of NH3 to grams. This is illustrated below:
Molar mass of NH3 = 14 + (3x1) = 17g/mol
Number of mole of NH3 = 5.02 moles
Mass of NH3 =..?
Mass = mole x molar Mass
Mass of NH3 = 5.02 x 17
Mass of NH3 = 85.34g
Therefore, 85.34g of NH3 is produced.
Answer:
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Explanation:
Here we have to calculate the amount of 
ion present in the sample.
In the sample solution 0.122g of ion is present.
The reaction happens on addition of excess BaCl₂ in a sample solution of potassium sulfate (K₂SO₄) and sodium sulfate [(Na)₂SO₄] can be written as-
K₂SO₄ = 2K⁺ +
(Na)₂SO₄=2Na⁺ +
Thus, BaCl₂+ = BaSO₄↓ + 2Cl⁻ .
(Na)₂SO₄ and K₂SO₄ is highly soluble in water and the precipitation or the filtrate is due to the BaSO₄ only. As a precipitation appears due to addition of excess BaCl₂ thus the total amount of ion is precipitated in this reaction.
The precipitate i.e. barium sulfate (BaSO₄)is formed in the reaction which have the mass 0.298g.
Now the molecular weight of BaSO₄ is 233.3 g/mol.
We know the molecular weight of sulfate ion () is 96.06 g/mol. Thus in 1 mole of BaSO₄ 96.06 g of ion is present.
Or. we may write in 233.3 g of BaSO₄ 96.06 g of ion is present. So in 1 g of BaSO₄ 
g of ion is present.
Or, in 0.298 g of the filtered mass (0.298×0.411)=0.122g of ion is present.
