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SVEN [57.7K]
3 years ago
12

Select all that apply.

Chemistry
2 answers:
PtichkaEL [24]3 years ago
8 0

Answer:

Nitrogen , phosphorus and potassium are obtained by plants from the soil.

Explanation:

N,P.K are the most important plant nutrients,

<h2><u>Nitrogen :</u></h2>
  1. Nitrogen helps above grounded safety growth

     2. It provide dark green color to leaves.

<h2><u>Phosphorus:</u></h2>
  1. Phosphorus also help root growth

     2.Protect the plant from diseases.

<h2><u>Potassium:</u></h2>
  1. Potassium is needed for making of chlorophyll.
prohojiy [21]3 years ago
5 0
My guess is the last 2... <span>phosphorus and potassium</span>
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3 0
3 years ago
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Question 6 The mineral barite, (BaSO.) has a ke of 1.1 x 10" at 25°C. Calculate the solubility of barium sulfate in water, in: 6
Sav [38]

Explanation:

(6.1).    The reaction equation will be as follows.

             BaSO_{4}(s) \rightleftharpoons Ba^{2+}(aq) + SO^{2-}_{4}(aq)

Assuming the value of K_{sp} as 1.1 \times 10^{-10} and let the solubility of each specie involved in this reaction is "s". The expression for K_{sp} will be as follows.

            K_{sp} = [Ba^{2+}][SO^{-}_{2}]    (Solids are nor considered)

                        = s \times s

                   s = \sqrt{K_{sp}}

                      = \sqrt{1.1 \times 10^{-10}}

                      = 1.05 \times 10^{-5}

Therefore, solubility of barium sulfate in water is 1.05 \times 10^{-5}.

(6.2).   As the molar mass of BaSO_{4} is 233.38 g/mol

Therefore, the solubility is g/L will be calculated as follows.

                233.38 g/mol \times 1.05 \times 10^{-5}

                  = 2.45 \times 10^{-3} g/L

Therefore, solubility of barium sulfate in grams per liter is 2.45 \times 10^{-3} g/L.

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3 years ago
True or False: The first part of an organism’s scientific name is its genus and the second part is its species.
castortr0y [4]

It is True :)

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4 0
4 years ago
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I believe it’s Hydrogen
5 0
4 years ago
At what temperature does uranium hexafluoride have a density of 0.5820 g/L at 0.5073 atm?
Svet_ta [14]

Answer:

4204 K

Explanation:

Step 1: Data

<em>Given data</em>

  • Density of uranium hexafluoride (ρ): 0.5820 g/L
  • Pressure of uranium hexafluoride (P): 0.5073 atm

<em>Required data</em>

  • Universal gas constant (R): 0.08206 atm.L/mol.K
  • Molar mass of uranium hexafluoride (M): 352.02 g/mol

Step 2: Calculate the temperature of the gas

We will use the following expression derived from the ideal gas equation.

P × M = ρ × R × T

T = P × M/ρ × R

T = 0.5073 atm × (352.02 g/mol)/(0.5820 g/L) × (0.08206 atm.L/mol.K)

T = 4204 K

8 0
3 years ago
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