Differentiate between sandy soil, clayey soil and loamy soil on the basis of the growth of plants????

1 answer:

8 0

The mixture of rock particle sand humus is called the soil.
If soil contains greater proportion of big particles it is called sandy soil. If the proportion of fine particles is relatively higher, then it is called clayey soil. If the amount of large and fine particles is about the same, then the soil is called loamy.

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Not yet answered Marked out of 2.00 What is the concentration of NH4+ in 60.0 mL of a 0.50 M solution of (NH4)3PO4? (To write yo

Softa [21]

Answer : The concentration of $NH_4^+$ ion is $0.15E^1M$

Explanation :

First we have to calculate the moles of $(NH_4)_3PO_4$ .

$\text{Moles of }(NH_4)_3PO_4=\text{Concentration of }(NH_4)_3PO_4\times \text{Volume of solution}=0.50M\times 0.06L=0.03mole$

The balanced chemical reaction will be:

$(NH_4)_3PO_4\rightleftharpoons 3NH_4^++PO_4^{3-}$

From the reaction we conclude that,

1 mole of $(NH_4)_3PO_4$ dissociate to give 3 moles of $NH_4^+$ ion and 1 mole of $PO_4^{3-}$ ion

So,

0.03 mole of $(NH_4)_3PO_4$ dissociate to give $3\times 0.03=0.09$ moles of $NH_4^+$ ion and 0.03 mole of $PO_4^{3-}$ ion

Now we have to calculate the concentration of $NH_4^+$ ion.

$\text{Concentration of }NH_4^+=\frac{\text{Moles of }NH_4^+}{\text{Total volume}}$

$\text{Concentration of }NH_4^+=\frac{0.09mole}{0.06L}=1.5M=0.15E^1M$

Therefore, the concentration of $NH_4^+$ ion is $0.15E^1M$

4 0

4 years ago

A student requires 2.00 L of 0.100 M NH4NO3 from a 1.75 M NH4NO3 stock solution. What is the correct way to get the solution?

EleoNora [17]

The student needs to dilute the stock to get the proper concentration. And during the dilution, the molar number of solute will not change. So the volume of stock needed is 2*0.1/1.75=0.114 L. Then dilute to 2.00 L.

3 0

4 years ago

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Select the correct answer.

natta225 [31]

I think the answer is b hope this helps

7 0

3 years ago

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2. If 4c-3= -31, what is the value of -2c+11

svetoff [14.1K]

Explanation: