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Feliz [49]
3 years ago
10

A multi-nutrient fertilizer contains several different nitrogen containing compounds. The fertilizer is 53.2% CH4N2O (urea), 24.

1% KNO3 , and 11.8% (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 2.20g N to a plant?
Chemistry
1 answer:
maks197457 [2]3 years ago
3 0

Answer:

7.19g of fertilizer are required to provide 2.20g of N to a plant

Explanation:

In 100g of fertilizer, the amount of nitrogen could be obtained as follows using the ratio of molar mass of nitrogen*atoms of nitrongen / molar mass of the molecule:

CH4N2O = 53.2g * (14g/mol*2 / 60.06g/mol) = 24.8gN

KNO3 = 24.1g * (14g/mol*1 / 101.10g/mol) = 3.3g N

(NH4)2HPO4 = 11.8g * (14g/mol*2 / 132.06g/mol) = 2.5g N

The mass of nitrogen in 100g of fertilizer is 24.8g + 3.3g + 2.5g = 30.6g N

That is: 30.6g N / 100g of fertilizer.

To obtain 2.20g of nitrogen are required:

2.20g N * (100g Fertilizer / 30.6g N) =

<h3>7.19g of fertilizer are required to provide 2.20g of N to a plant</h3>
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In general, in what type of solvent (non-polar, moderately polar, or highly polar) are polar solutes most soluble? Explain why.
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AveGali [126]

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Determine the volume occupied by 0.352 mole of a gas at 25⁰C if the pressure is 81.8 kPa.
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Data:
p (pressure) = 81.8 kPa = 81.8*10³ Pa ≈ 8.07 atm
v (volume) = ? (in L)
n (number of mols) = 0.352 mol
R (Gas constant) = 0.082 (atm*L/mol*K)
T (temperature) = 25ºC converting to Kelvin, we have:
TK = TC + 273 → TK = 25 + 273 → TK = 298

Formula:
p*V =n*R*T

Solving:
p*V =n*R*T
8.07*V = 0.352*0.082*298
8.07V \approx 8.60
V \approx  \frac{8.60}{8.07}
\boxed{\boxed{V \approx 1.06\:L}}
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