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A fertilizer contains \(21 \%\) nitrogen by mass. What mass of this fertilizer, in kilograms, is required for an application requiring \(225 \mathrm{g}\) of nitrogen?

Short Answer

Expert verified
The mass of fertilizer required for an application requiring 225 g of nitrogen is approximately 1.07 Kg.

Step by step solution

01

Convert Mass of Desired Nitrogen to Kilograms

225 g of nitrogen is equal to \(225 \div 1000 = 0.225\) Kg.
02

Calculate Required Mass of Fertilizer Using Proportions

Given that the fertilizer contains 21% nitrogen by mass, this means that if you have 1kg of the fertilizer, the mass of nitrogen in it would be 0.21Kg (since 21% of 1kg equals 0.21kg). So, to find the mass of the fertilizer that gives 0.225kg of nitrogen, set up a proportion and solve for x (mass of fertilizer). Thus: \( \frac{21}{100} : 1 = 0.225 : x \) , solving for x, yields \( x = \frac{1 \times 0.225}{0.21} \)
03

Final Result

After solving the proportion for x, the desired mass of fertilizer in kilograms is obtained.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Proportions in Chemistry
Proportions in chemistry are crucial for solving many problems, especially when dealing with mixtures and solutions. Proportions help us understand the relationship between different components in a chemical mixture. We use them to ensure the right amounts of each element are present, leading to the desired chemical reaction or effect.

In this exercise, to find how much of a certain fertilizer is needed when you know it contains a specific percentage of nitrogen, you set up a proportion. A proportion is essentially a statement that two ratios are equal.

For example, if you know that 1 kilogram of a fertilizer contains 21% nitrogen by mass, you can write this as a ratio: \( \frac{21}{100} \). To find out how much of the fertilizer provides 0.225 kilograms of nitrogen, you set a proportion that connects these two scenarios. Solving proportions lets you find unknown quantities when you know the ratios involved.
Examining Fertilizer Composition
Fertilizer composition refers to the elements or nutrients that make up a fertilizer. These compositions are often presented in percentages by mass for key nutrients like nitrogen (N), phosphorus (P), and potassium (K), usually in the form of an "N-P-K" ratio.

This labeling standard makes it easy to understand what percentage of each nutrient is in the fertilizer. For instance:
  • An "N-P-K" ratio of 21-0-0 means 21% nitrogen, 0% phosphorus, and 0% potassium.
In our exercise, we analyzed a fertilizer with 21% nitrogen. Understanding this percentage helps determine how much fertilizer is necessary to provide a specific amount of nitrogen.

When choosing a fertilizer, knowing its composition is vital for fulfilling the specific nutrient requirements of plants, ensuring healthy growth.
Calculating Nitrogen Percentage
Calculating the nitrogen percentage within a fertilizer involves using given data to understand how much nitrogen is present in a total mass.

In our problem, we knew that 21% of the fertilizer’s mass is nitrogen. To find out how much fertilizer you need for a certain amount of nitrogen, knowing how to manipulate percentages and convert between units is important.

Steps to calculate this include:
  • Understanding that 21% nitrogen means every 100 kilograms of fertilizer contains 21 kilograms of nitrogen.
  • Converting any given nitrogen requirements from grams to kilograms as needed (since percentages are often based on kilograms).
  • Setting up the proportion to solve for the total fertilizer mass required to meet the nitrogen needs (using the equation \( x = \frac{1 \times 0.225}{0.21} \)).
This calculation ensures the correct amount of nutrient is added without excess, enabling optimal plant growth.

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Most popular questions from this chapter

Indicate whether each of the following is an exact number or a measured quantity subject to uncertainty. (a) the number of pages in this text (b) the number of days in the month of January (c) the area of a city lot (d) the distance between the centers of the atoms in a gold medal

Of the following masses, two are expressed to the nearest milligram. The two are (a) \(32.7 \mathrm{g}\); (b) \(0.03271 \mathrm{kg} ;(\mathrm{c}) 32.7068 \mathrm{g} ;(\mathrm{d}) 32.707 \mathrm{g} ;(\mathrm{e}) 30.7 \mathrm{mg};\) (f) \(3 \times 10^{3} \mu g.\)

The fact that the volume of a fixed amount of gas at a fixed temperature is inversely proportional to the gas pressure is an example of (a) a hypothesis; (b) a theory; (c) a paradigm; (d) the absolute truth; (e) a natural law.

A lump of pure copper weighs \(25.305 \mathrm{g}\) in air and 22.486 g when submerged in water \((d=0.9982 \mathrm{g} / \mathrm{mL})\) at \(20.0^{\circ} \mathrm{C} .\) Suppose the copper is then rolled into a \(248 \mathrm{cm}^{2}\) foil of uniform thickness. What will this thickness be, in millimeters?

A technique once used by geologists to measure the density of a mineral is to mix two dense liquids in such proportions that the mineral grains just float. When a sample of the mixture in which the mineral calcite just floats is put in a special density bottle, the weight is 15.4448 g. When empty, the bottle weighs 12.4631 g, and when filled with water, it weighs 13.5441 g. What is the density of the calcite sample? (All measurements were carried out at \(25^{\circ} \mathrm{C}\), and the density of water at \(25^{\circ} \mathrm{C}\) is \(0.9970 \mathrm{g} / \mathrm{mL}\) ). At the left, grains of the mineral calcite float on the surface of the liquid bromoform \((d=2.890 \mathrm{g} / \mathrm{mL})\) At the right, the grains sink to the bottom of liquid chloroform \((d=1.444 \mathrm{g} / \mathrm{mL}) .\) By mixing bromoform and chloroform in just the proportions required so that the grains barely float, the density of the calcite can be determined (Exercise 62).

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