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Katen [24]
1 year ago
10

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

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Rank the compounds by the ease with which they ionize under sn1 conditions. rank the compounds from easiest to hardest. to rank
marysya [2.9K]
The rate of Formation of Carbocation mainly depends on two factors'

                    1)  Stability of Carbocation:
                                                              The ease of formation of Carbocation mainly depends upon the ionization of substrate. If the forming carbocation id tertiary then it is more stable and hence readily formed as compared to secondary and primary.

                     2) Ease of detaching of Leaving Group:
                                                                                   The more readily and easily the leaving group leaves the more readily the carbocation is formed and vice versa. In given scenario the carbocation formed is tertiary in all three cases, the difference comes in the leaving group. So, among these three substrates the one containing Iodo group will easily dissociate to form tertiary carbocation because due to its large size Iodine easily leaves the substrate, secondly Chlorine is a good leaving group compared to Fluoride. Hence the order of rate of formation of carbocation is,

                                            R-I > R-Cl > R-F

                                               B   >  C  >  A

3 0
1 year ago
A gas that has a volume of 28 liters, a temperature of 45C, And an unknown pressure has its volume increased to 34 liters and it
patriot [66]

Answer:

P1 = 2.5ATM

Explanation:

V1 = 28L

T1 = 45°C = (45 + 273.15)K = 318.15K

V2 = 34L

T2 = 35°C = (35 + 273.15)K = 308.15K

P1 = ?

P2 = 2ATM

applying combined gas equation,

P1V1 / T1 = P2V2 / T2

P1*V1*T2 = P2*V2*T1

Solving for P1

P1 = P2*V2*T1 / V1*T2

P1 = (2.0 * 34 * 318.15) / (28 * 308.15)

P1 = 21634.2 / 8628.2

P1 = 2.5ATM

The initial pressure was 2.5ATM

3 0
1 year ago
An old sample of concentrated sulfuric acid to be used in the laboratory is approximately 98.1 percent h2so4 by mass. calculate
Airida [17]
Basis: 100 mL solution

From the given density, we calculate for the mass of the solution.

                        density  = mass / volume
                        mass = density x volume

                       mass = (1.83 g/mL) x (100 mL) = 183 grams

Then, we calculate for the mass H2SO4 given the percentage.
                   
                      mass of H2SO4 = (183 grams) x (0.981) = 179.523 grams

Calculate for the number of moles of H2SO4,
                  moles H2SO4 = (179.523 grams) / (98.079 g/mol)
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Molarity:
                     M = moles H2SO4 / volume solution (in L)
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Molality:
                     m = moles of H2SO4 / kg of solvent
                        = 1.83 moles / (183 g)(1-0.983)(1 kg/ 1000 g) = 588.24 m
8 0
2 years ago
Write an equation that represents the action in water of rubidium hydroxide as an Arrhenius base.
Anika [276]

Answer:

RbOH  → Rb⁺ +  OH⁻

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Explanation:

Arrhenius theory states that a compound is considered a base, if the compound can generate OH⁻ ions in aqueous solution.

Our compound is the RbOH.

When it is put in water, i can dissociate like this:

RbOH  → Rb⁺ +  OH⁻

As the hydroxide can gives the OH⁻ in water, it is considered as an Arrhenius's base

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Answer:

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Explanation:

4 0
2 years ago
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