Rank the following fertilizers in decreasing order of mass percentage of nitrogen: (nh4)2hpo4 (nh4)2so4 kno3 (nh4)h2po4 nh3 nh4no3

        NH₃ > NH₄NO₃ > (NH₄)₂HPO₄ > (NH₄)₂SO₄ > KNO₃ > (NH₄)H₂PO₄

Soution:

In (NH₄)₂HPO₄:

Mass of Nitrogen  =  N × 2  =  14 × 2  =  28 g.mol⁻¹

Molar Mass of (NH₄)₂HPO₄  =  132.06 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of (NH₄)₂HPO₄ × 100

Mass %age  =  28 g.mol⁻¹ / 132.06 g.mol⁻¹ × 100

Mass %age  =  21.20 %

In (NH₄)₂SO₄:

Mass of Nitrogen  =  N × 2  =  14 × 2  =  28 g.mol⁻¹

Molar Mass of (NH₄)₂SO₄  =  132.14 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of (NH₄)₂SO₄ × 100

Mass %age  =  28 g.mol⁻¹ / 132.14 g.mol⁻¹ × 100

Mass %age  =  21.18 %

In KNO₃:

Mass of Nitrogen  =  N × 1  =  14 × 1  =  14 g.mol⁻¹

Molar Mass of KNO₃  =  101.10 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of KNO₃ × 100

Mass %age  =  14 g.mol⁻¹ / 101.10 g.mol⁻¹ × 100

Mass %age  =  13.84 %

In (NH₄)H₂PO₄:

Mass of Nitrogen  =  N × 1  =  14 × 1  =  14 g.mol⁻¹

Molar Mass of (NH₄)H₂PO₄  =  115.03 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of (NH₄)H₂PO₄ × 100

Mass %age  =  14 g.mol⁻¹ / 115.03 g.mol⁻¹ × 100

Mass %age  =  12.17 %

In NH₃:

Mass of Nitrogen  =  N × 1  =  14 × 1  =  14 g.mol⁻¹

Molar Mass of NH₃  =  132.14 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of NH₃ × 100

Mass %age  =  14 g.mol⁻¹ / 17.03 g.mol⁻¹ × 100

Mass %age  =  82.20 %

In NH₄NO₃:

Mass of Nitrogen  =  N × 2  =  14 × 2  =  28 g.mol⁻¹

Molar Mass of NH₄NO₃  =  80.04 g.mol⁻¹

Mass %age  =  Mass of N / M.Mass of NH₄NO₃ × 100

Mass %age  =  28 g.mol⁻¹ / 80.04 g.mol⁻¹ × 100

Mass %age  =  34.98 %

Earth's rotation cause the global convention current between the equator and the poles

For cyclohexane compounds drawn in a chair conformation:   -axial bonds are vertical  -equatorial bonds are parallel to two bonds within the ring.  thus, as i assume the chair conformation is drawn:   -the x group is in an axial position, as are the five other vertical bonds.  -three axial bonds are on the top face of the ring  -the other three axial bonds are on the bottom face  the position of the second substituent on each ring carbon is equatorial.  gauche groups must be on an adjacent carbon such as carbon 6 and carbon 2, and also gauche groups are rotated by 60 degrees relative to each other.  drawing the newman projections for c1–c2 and c1–c6 aids in visualizing the groups that are gauche to the y group on c1.  click this link and watch some academic videos that may you more understand the topic