Answer: 9.08 L
Explanation:
To calculate the moles :

According to stoichiometry :
4 moles of
require = 3 moles of 
Thus 0.54 moles of
will require=
of 
Standard condition of temperature (STP) is 273 K and atmospheric pressure is 1 atm respectively.
According to the ideal gas equation:

P = Pressure of the gas = 1 atm
V= Volume of the gas = ?
T= Temperature of the gas = 273 K
R= Gas constant = 0.0821 atmL/K mol
n= moles of gas= 0.405

Thus 9.08 L of
at STP would be required
Organism<span>. ' s rate of mutation is directly proportional to its adaptability. .... D. The cell membrane contains </span>genetic<span> information of the cell. .... </span>What<span> cellular structure was the dialysis tubing most likely</span>representing<span> in this experiment? ...... C </span>Rr<span> and </span>rr<span> only ... A healthy </span>individual<span> is a carrier of a lethal allele but is unaffected by it.</span>
<u>Answer:</u> The mass of iron (III) nitrate is 11.16 g/mol
<u>Explanation:</u>
To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

We are given:
Molarity of solution = 0.3556 M
Molar mass of Iron (III) nitrate = 241.86 g/mol
Volume of solution = 129.8 mL
Putting values in above equation, we get:

Hence, the mass of iron (III) nitrate is 11.16 g/mol
They were natural philosophers that used experimentation to establish the cause effect relations and explanations, in an epoque when many "theories' were not supported by experiments but by some authority (antique philosophers like Aristoteles or the Catholic Church).
Answer : The concentration of
needed is, 
Explanation :
First we have to calculate the mole of phosphate.
As we are given that, 1 mg P/L that means, 1 mg of phosphate present in 1 L of solution.

Molar mass of phosphate = 94.97 g/mole

Now we have to calculate the concentration of phosphate.


Now we have to calculate the concentration of
.
The second equilibrium reaction is,

The solubility constant expression for this reaction is:
![K_{sp}=[Fe^{3+}][PO_4^{3-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BFe%5E%7B3%2B%7D%5D%5BPO_4%5E%7B3-%7D%5D)
Given: 
![\frac{1}{4}=[Fe^{3+}]\times 1.053\times 10^{-5}mol/L](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B4%7D%3D%5BFe%5E%7B3%2B%7D%5D%5Ctimes%201.053%5Ctimes%2010%5E%7B-5%7Dmol%2FL)
![[Fe^{3+}]=2.37\times 10^4M](https://tex.z-dn.net/?f=%5BFe%5E%7B3%2B%7D%5D%3D2.37%5Ctimes%2010%5E4M)
Thus, the concentration of
needed is, 