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3 years ago

Phosphorus is:

Chemistry

2 answers:

docker41 [41]3 years ago

4 0

its returned to the soil by rhizobium

Readme [11.4K]3 years ago

3 0

Answer:

added to the soil through weathering

Explanation:

1) found free in the atmosphere : in it is found in the form of phosphates etc.

2 returned to the soil by rhizobium : Rhizobium is a bacteria which is involved in nitrogen fixation.

3 added to the soil through weathering : It is added to soil again by weathering and is absorbed by plants in the form of phosphates.

4 usually very abundant in an ecosystem: it is found in earth nearly 0.1% by weight.

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Convert 4.1x10^-4 moles of carbon into atoms.

Ratling [72]

Answer:

2.5×10^20atoms

Explanation:

number of moles= number of particles/Avogadro's constant

number of particles=Avogadro's constant × number of moles

number of particles=(6.02×10^23) × (4.1×10^-4)

number of particles=2.5×10^20

8 0

3 years ago

What is the percent by mass of water in MgSO4 • 7H2O? A. 51.1% B. 195% C. 56.0% D. 21.0%

lbvjy [14]

Answer : The correct option is, (A) 51.1%

Explanation :

Mass percent : It is defined as the mass of the given component present in the total mass of the compound.

Formula used :

$\text{Mass} \%H_2O=\frac{\text{Mass of }H_2O}{\text{Mass of }MgSO_4.7H_2O}\times 100$

First we have to calculate the mass of $H_2O$ and $MgSO_4.7H_2O$ .

Mass of $H_2O$ = 18 g/mole

Mass of $7H_2O$ = 7 × 18 g/mole = 126 g/mole

Mass of $MgSO_4.7H_2O$ = 246.47 g/mole

Now put all the given values in the above formula, we get the mass percent of $H_2O$ in $MgSO_4.7H_2O$ .

$\text{Mass} \%H_2O=\frac{126g/mole}{246.47g/mole}\times 100=51.1$

Therefore, the mass percent of $H_2O$ in $MgSO_4.7H_2O$ is, 51.1%

6 0

3 years ago

A gas occupies 3.00 L at standard temperature. The volume at 350 C is 6.84 L.<br> True <br> False

kupik [55]

Answer:

TRUE

Explanation:

Standard temperature is 273K and 350°C is 623K.

V/V' = T/T'

3/6.84 = 273/T'

T' = (273×6.84)/3

=> T' = 623K

So it is true.

7 0

3 years ago

Which chemical reaction involves the fewest oxygen atoms?

marin [14]

Answer:

Explanation:

do you know what answer choice is correct please is need the answer

5 0

3 years ago

Can somebody help me. Will mark brainliest.

pickupchik [31]

Answer:

2.27%

Explanation:

The observed value of the boiling point = 71.8°C

The actual value of the boiling point = 70.2°C

We need to find the percentage error of the student's measurement. The percentage error in any value is given by :

$\%=\dfrac{\text{calculated value-actual value}}{\text{actual value}}\times 100\\\\=\dfrac{71.8-70.2}{70.2}\times 100\\\\=2.27\%$

So, the required percentage error is 2.27%.

8 0

3 years ago