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andrezito [222]
3 years ago
9

Can the same fertilizer be applied to all types of soil ? Explain​

Chemistry
1 answer:
Arte-miy333 [17]3 years ago
4 0
You cant use the same fertilizer for all soils bc differnt soils have different nutrients like lets say ur growing tomatoes with cabbage it wouldnt work bc they need different nutrients for them to grow. This is used with trees, veggies, plants and fruits. it is true somw can use the same fertilizer if they can be grown with it. if you give it the wrong nutrients it can die or not grow
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The substance chlorine monoxide, ClO(g), is important in atmospheric processes that lead to the depletion of the ozone layer. Th
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Answer:

B = Based on the electronegativities of the elements, which atom would you expect to have a partial negative charge in the ClO molecule?

Atom O or Atom Cl

Explanation:

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. The element with the following
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Explanation:

(a) Chlorine the first row Is 1s the second is 2s followed by 2p...etc as Chlorine has 17 electrons

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What do the following have in common: MgCl2, AlF3, CaI2, KCl
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Answer:

They are salts, ionic compounds

Explanation:

To know if they are covalent or ionic we need to check their difference in electronegativity. As an example mgcl2, mg has electronegativity of 1.2 and cl 3 so 3-1.2=1.8 which is bigger than 1.7 means it's ionic therefore they are salts

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Which soil layer is the least similar to the underlying rock
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Using henry's law, calculate the molar concentration of o2 in the surface water of a mountain lake saturated with air at 20 ∘c a
s2008m [1.1K]

Answer: The molar concentration of oxygen gas in water is 1.43\times 10^{-7} mol/L.

Explanation:

Partial pressure of the O_2gas = 685 torr = 0.8905 bar

1 torr = 0.0013 bar

According Henry's law:

p_{o_2}=K_H\times\chi_{O_2}

Value of Henry's constant of oxygen gas at 20 °C in water = 34860 bar

0.0013=34860 bar\times \chi_{O_2}

\chi_{O_2}=\frac{0.8905 bar}{34680 bar}=2.56\times 10^{-5}

Let the number of moles of O_2 gas in 1 liter water be n.

1 Liter water = 1000 g of water

Moles of water in 1 L n_w=\frac{1000 g}{18 g/mol}=55.55 mol

\chi_{O_2}=\frac{n}{n+n_w}

2.56\times 10^{-5}=\frac{n}{n+55.55}

n=1.43\times 10^{-7} moles

Molarity=\frac{\text{Moles of}O_2}{Volume}

Molar concentration of oxygen gas in 1 L of water:

=\frac{1.43\times 10^{-7} moles}{1 L}=1.43\times 10^{-7} mol/L

The molar concentration of oxygen gas in water is 1.43\times 10^{-7} mol/L.

4 0
3 years ago
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