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jenyasd209 [6]
2 years ago
15

To help farmers and gardeners, commercial fertilizers have a big three-number "NPK" label on the bag that gives the amounts of t

hree key plant nutrients: nitrogen, phosphorus and potassium. For example, a bag of fertilizer has by mass nitrogen, by mass potassium as potash , and by mass phosphorus as phosphate . Suppose a certain fertilizer has the following composition: ingredient percent by mass ammonium phosphate calcium nitrate urea inert ingredients What should the first (nitrogen) number on the bag label be? That is, calculate the percent by mass of nitrogen in this fertilizer. Round your answer to the nearest percent.

Chemistry
1 answer:
Ludmilka [50]2 years ago
8 0

Answer:

The number on the lag label should be 15.

Explanation:

It seems your question is incomplete, as it is lacking the working values. An internet search showed me the full question, you can see it in the attached picture.

Let's say we have 100 g of the fertilizer.

  • <em>45 g are of ammonium phosphate</em> ( (NH₄)₃PO₄ ), of which:
  • 45 g (NH₄)₃PO₄ * \frac{42 g N}{149g(NH_{4})_{3}PO_{4}} = 12.7 g are of Nitrogen.

(We used the molar mass of ammonium phosphate in the denominator and three times the molar mass of nitrogen in the numerator)

  • <em>18 g are of calcium nitrate</em> (Ca(NO₃)₂), of which:
  • 16 g Ca(NO₃)₂ *\frac{28gN}{164gCa(NO_{3})_{2}} = 2.73 g are of Nitrogen.

So in total there are (12.7+2.73) 15.43 g of Nitrogen in 100 g of the fertilizer. So the percent by mass of nitrogen is 15.43%.

Rounding to the nearest percent the answer is 15.

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Calculate the number of kilojoules of energy required to convert 50.0 grams of solid DMSO initially at a temperature of 19.0°C t
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Answer:

20.79 kilojoules

Explanation:

Using Q = m×c×∆T

Where;

Q = Quantity of heat (J)

c = specific heat capacity of solid DMSO (1.80 J/g°C)

m = mass of DMSO

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According to the provided information, m= 50g, initial temperature = 19.0°C, final temperature= 250.0°C

Q = m×c×∆T

Q = 50 × 1.80 × (250°C - 19°C)

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To convert Joules to kilojoules, we divide by 1000 i.e.

20790/1000

= 20.79 kilojoules

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1) How many aluminum atoms are there in 3.50 grams of Al2O3?
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Answer:

Aluminium atoms =  4.13 *10^22 aluminium atoms

The correct answer is E

Explanation:

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Mass of Al2O3 = 3.50 grams

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Number of Avogadro = 6.022 * 10^23 /mol

Step 2: Calculate moles Al2O3

Moles Al2O3 = mass Al2O3 / molar mass Al2O3

Moles Al2O3 = 3.50 grams / 101.96 g/mol

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Step 3: Calculate moles Aluminium

In 1 mol Al2O3 we have 2 moles Al

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Step 4: Calculate aluminium atoms

Aluminium atoms =  moles aluminium * Number of Avogadro

Aluminium atoms =  0.0686 * 6.022 * 10^23

Aluminium atoms =  4.13 *10^22 aluminium atoms

The correct answer is E

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