All categories

1 year ago

A bag of fertilizer is labeled 10-20-20. What is the actual percentage of phosphorus in the fertilizer?

Chemistry

1 answer:

aleksandr82 [10.1K]1 year ago

8 0

Answer:

20% of phosphorus

Explanation:

A fertilizer is used to improve the fertility of soils. Most fertilizers contains the element nitrogen, phosphorus and potassium.

They are often designated NPK fertilizers.

Now we know that the numbers 10-20-20 depicts the nitrogen-phosphorus and potassium content of the fertilizer.

From the designation,

The actual percentage is 20% of phosphorus.

10% of nitrogen

20% of potassium

You might be interested in

Click the "draw structure" button to launch the drawing utility. Propane (CH3CH2CH3) has both 1° and 2° hydrogens. Draw the carb

OlgaM077 [116]

Answer:

a) find attached image 1

b) find attached image 2

Explanation :

The more stable radical is formed by a reaction with smaller bond dissociation energy.

since the bond dissociation for cleavage of the bond to form primary free radical is higher, more energy must be added to form it. This makes primary free radical higher in energy and therefore less stable than secondary free radical.

6 0

1 year ago

An ice cube with a volume of 45.0ml and a density of 0.9000g/cm3 floats in a liquid with a density of 1.36g/ml. what volume of t

Grace [21]

Answer : The volume of the cube submerged in the liquid is, 29.8 mL

Explanation :

First we have to determine the mass of ice.

Formula used :

$\text{Mass of ice}=\text{Density of ice}\times \text{Volume of ice}$

Given:

Density of ice = $0.9000g/cm^3=0.9000g/mL$

Volume of ice = 45.0 mL

$\text{Mass of ice}=0.9000g/mL\times 45.0mL$

$\text{Mass of ice}=40.5g$

The cube will float when 40.5 g of liquid is displaced.

Now we have to determine the volume of the cube is submerged in the liquid.

$\text{Volume of ice}=\frac{\text{Mass of liquid}}{\text{Density of liquid}}$

$\text{Volume of ice}=\frac{40.5g}{1.36g/mL}$

$\text{Volume of ice}=29.8mL$

Thus, the volume of the cube submerged in the liquid is, 29.8 mL

5 0

1 year ago

Suppose you are titrating vinegar, which is an acetic acid solution of unknown strength, with a sodium hydroxide solution accord

Marina CMI [18]

Answer:

$M_{acid}=0.563M$

Explanation:

Hello there!

In this case, given the neutralization of the acetic acid as a weak one with sodium hydroxide as a strong base, we can see how the moles of the both of them are the same at the equivalence point; thus, it is possible to write:

$M_{acid}V_{acid}=M_{base}V_{base}$

Thus, we solve for the molarity of the acid to obtain:

$M_{acid}=\frac{M_{base}V_{base}}{V_{acid}} \\\\ M_{acid}=\frac{33.98mL*0.1656M}{10.0mL}\\\\ M_{acid}=0.563M$

Regards!

5 0

1 year ago

Determine the number of moles and mass requested for each reaction in Exercise 4.42.

suter [353]

Answer:

(a) 0.22 mol Cl₂ and 15.4g Cl₂

(b) 2.89.10⁻³ mol O₂ and 0.092g O₂

(d) 1,666 mol CO₂ and 73,333 g CO₂

(e) 18.87 CuCO₃ and 2,330g CuCO₃

Explanation:

In most stoichiometry problems there are a few steps that we always need to follow.

Step 1: Write the balanced equation
Step 2: Establish the theoretical relationship between the kind of information we have and the one we are looking for. Those relationships can be found in the balanced equation.
Step 3: Apply conversion factor/s to the data provided in the task based on the relationships we found in the previous step.

(a)

Step 1:

2 Na + Cl₂ ⇄ 2 NaCl

Step 2:

In the balanced equation there are 2 moles of Na, thus 2 x 23g = 46g of Na. 46g of Na react with 1 mol of Cl₂. Since the molar mass of Cl₂ is 71g/mol, then 46g of Na react with 71g of Cl₂.

Step 3:

$10.0gNa.\frac{1molCl_{2} }{46gNa} =0.22molCl_{2}$

$10.0gNa.\frac{71gCl_{2}}{46gNa} =15.4gCl_{2}$

(b)

Step 1:

HgO ⇄ Hg + 0.5 O₂

Step 2:

216.5g of HgO form 0.5 moles of O₂. 216.5g of HgO form 16g of O₂.

Step 3:

$1.252gHgO.\frac{0.5molO_{2}}{216.5gHgO} =2.89.10^{-3} molO_{2}$

$1.252gHgO.\frac{16gO_{2}}{216.5gHgO} =0.092gO_{2}$

(c)

Step 1:

NaNO₃ ⇄ NaNO₂ + 0.5 O₂

Step 2:

16g of O₂ come from 1 mol of NaNO₃. 16g of O₂ come from 85g of NaNO₃.

Step 3:

$128gO_{2}.\frac{1molNaNO_{3}}{16gO_{2}} =8mol NaNO_{3}$

$128gO_{2}.\frac{85gNaNO_{3}}{16gO_{2}} =680gNaNO_{3}$

(d)

Step 1:

C + O₂ ⇄ CO₂

Step 2:

12 g of C form 1 mol of CO₂. 12 g of C form 44g of CO₂.

Step 3:

$20.0kgC.\frac{1,000gC}{1kgC} .\frac{1molCO_{2}}{12gC} =1,666molCO_{2$

$[tex]20.0kgC.\frac{1,000gC}{1kgC} .\frac{44gCO_{2}}{12gC} =73,333gCO_{2$ [/tex]

(e)

Step 1:

CuCO₃ ⇄ CuO + CO₂

Step 2:

79.5g of CuO come from 1 mol of CuCO₃. 79.5g of CuO come from 123.5g of CuCO₃.

Step 3:

$1.500kgCuO.\frac{1,000gCuO}{1kgCuO} .\frac{1mol CuCO_{3}}{79.5gCuO} =18.87molCuCO_{3}\\ 1.500kgCuO.\frac{1,000gCuO}{1kgCuO} .\frac{123.5g CuCO_{3}}{79.5gCuO} =2,330gCuCO_{3}$

5 0

1 year ago

what does the number of the places after the decimal in a measurement tell you about the precision of the instrument that record

snow_lady [41]

The more numbers after the decimal point there are, the more precise the instrument which recorded it is. For example, if one instrument during seismic activity records that the magnitude of the earthquake was 2.3, and another instrument recorded that it was 2.3645, the second instrument would have shown to be more precise.

3 0

1 year ago