Amulti-nutrient fertilizer contains several different nitrogen containing compounds. the fertilizer is 54.8% ch4n2o (urea), 26.3% kno3 , and 14.1% (nh4)2hpo4 by mass. the remainder of the fertilizer consists of substances that do not contain nitrogen. how much fertilizer should someone apply to provide 2.90 g n to a plant?

9.01 g of fertilizer

Explanation:

Assume initially that the amount in grams of the multinutrient fertilizer is 100 g, and calculate the amount of nitrogen supplied by every compound present in these 100 g of fertilizer.

1) Urea (CH₄N₂O):

Molar mass of urea: 60.06 g/molAtomic mass of N: 14.007 g/molTotal mass of N in the formula: 2 × 14.007 g/mol = 28.014 g/mol Amount of N in 100 g of compound: 100 g × 54.8% × 28.014 g / 60.06 g =  25.56 g

2) KNO₃

Molar mass of KNO₃: 101.1032 g/molAtomic mass of N: 14.007 g/ molTotal mass of N in the formula: 14.007 g/molAmount of N in 100 of the compound: 100 g × 26.3% × 14.007 / 101.1032 = 3.64 g

3) (NH₄)₂PO₄

Molar mass of (NH₄)₂PO₄: 132.06 g/molAtomic mass of N: 14.007 g/ molTotal mass of N in the formula: 2×14.007 g/mol = 28.014 g/molAmount of N in 100 of the compound: 100 g × 14.1% × 14.007 / 132.06 = 2.99 g

4) Total mass of N in 100 g of fertilizer:

Add all the amounts of N obtained above

25.56g + 3.69 g + 2.99 g = 32.19 g of N

5) Mass of fertilizer that should be applied to provide 2.90 g of N to a plant.

Set a proportion:

32.19 g of N / 100 g of fertilizer = 2.90 g of N / X

Solve for X:

X = 2.90 g of N × 100 g of fertilizer / 32.19 g of N = 9.01 g of fertilizer

That is the 9.01 g of fertilizer should be applied to provide 2.90g  of N to a plant.

False because saturated fatty acids contain only single bond

All the elements are arranged by melting points on the periodic table. in row they go from metals to gasses. in the row from sodium to argon, sodium is more reactive than the one on its right, magnesium.