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.
1 answer:
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 *
= 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 *
= 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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Answer:
Explanation:
Hello,
In this case, we can compute the change in the solution enthalpy by using the following formula:
Whereas the mass of the solution is 350 g, the specific heat capacity is 4.184 J/g °C and the change in the temperature is 1.34 °C, therefore, we obtain:
It is important to notice that the mass is just 350 g that is the reacting amount and by means of the law of the conservation of mass, the total mass will remain constant, for that reason we compute the change in the enthalpy as shown above, which is positive due to the temperature raise.
Best regards.
Answer:
2 water + sugar + lemon juice → 4 lemonade
Moles of water present in 946.36 g of water=\frac{946.36 g}{236.59 g/mol}=4 mol=
236.59g/mol
946.36g
=4mol
Moles of sugar present in 196.86 g of water=\frac{196.86 g}{225 g/mol}=0.8749 mol=
225g/mol
196.86g
=0.8749mol
Moles of lemon juice present in 193.37 g of water=\frac{193.37 g}{257.83 g/mol}=0.7499 mol=
257.83g/mol
193.37g
=0.7499mol
Moles of lemonade in 2050.25 g of water=\frac{2050.25 g}{719.42 g/mol}=2.8498 mol=
719.42g/mol
2050.25g
=2.8498mol
As we can see that number of moles of lemon juice are limited.
So, we will consider the reaction will complete in accordance with moles of lemon juice.
1 mole lemon juice reacts with 2 mol of water,then 0.7499 mol of lemon juice will react with:
\frac{2}{1}\times 0.7499 mol = 1.4998 mol
1
2
×0.7499mol=1.4998mol of water
Mass of water used = 1.4998 mol × 236.59 g/mol=354.8376 g
Water remained unused = 946.36 g - 354.8376 g =591.5223 g
1 mole lemon juice reacts with mol of sugar,then 0.7499 mol of lemon juice will react with:
\frac{1}{1}\times 0.7499 mol = 0.7499 mol
1
1
×0.7499mol=0.7499mol of water
Mass of sugar used = 0.7499 mol × 225 g/mol = 168.7275 g
Sugar remained unused = 196.86 g - 28.1325 g
1 mole of lemon juice gives 4 moles of lemonade.
Then 0.7499 mol of lemon juice will give:
\frac{4}{1}\times 0.7499 mol=2.996 mol
1
4
×0.7499mol=2.996mol of lemonade
Mass of lemonade obtained = 2.996 mol × 719.42 g/mol = 2157.9722 g
Theoretical yield of lemonade = 2157.9722 g
Experimental yield of lemonade = 2050.25 g
Percentage yield of lemonade:
\frac{\text{Experimental yield}}{\text{theoretical yield}}\times 100
theoretical yield
Experimental yield
×100
\frac{2050.25 g}{2157.9722 g}\times 100=95.00\%
2157.9722g
2050.25g
×100=95.00%
