Astudent produces 1.34g of mgnh4po4*6h2o (mm =245.1 g/mol) from a sample of fertilizer, how grams of phosphorous were in the fertilizer?

of phosphorus is present in 1.34 g of fertilizer.

Further Explanation:

Mole is a measure of the amount of substance. It is defined as the mass of a substance that has the same number of fundamental units as there are atoms in 12 g of carbon-12. Such fundamental units can be atoms, molecules or formula units.

The formula to calculate the moles of fertilizer is as follows:

                …… (1)

The given mass of is 1.34 g.

The molar mass of is 245.1 g/mol.

Substitute these values in equation (1)

The chemical formula of the fertilizer is . This indicates that one mole of the fertilizer contains one mole of phosphorus.

So the number of moles of phosphorus in 0.0054671563 moles of fertilizer can be calculated as follows:

The formula to calculate the mass of phosphorus is as follows:

…… (2)

The number of moles of phosphorus is 0.0054671563 mol.

The molar mass of phosphorus is 30.97 g/mol.

Substitute these values in equation (2).

So, 0.17 grams of phosphorus is present in the fertilizer.

Learn more:

1. Calculate the moles of chlorine:

2. What is the number of moles of ions in HCl solution?

Answer details:

Grade: High School

Subject: Chemistry

Chapter: Mole concept

Keywords: moles, fertilizer, 0.17 g, MgNH4PO4.6H2O, molar mass, phosphorus, moles, mole, molar mass, 30.97 g/mol and0.0054671563 mol.

8

explanation:

each principal energy level above the first contains one s orbital and three p orbitals. a set of three p orbitals, called the p sublevel, can hold a maximum of six electrons. therefore, the second level can contain a maximum of eight electrons - that is, two in the s orbital and 6 in the three p orbitals.

mark brainliest and have a great day!

i think it is true since you can test the answer and can use observations to complete it.