1.1943571. The molar mass of K3PO4 is 212.27. After using stochieometry, you get your answer.
Answer:
![[Fe^{+3}]=0.700 M](https://tex.z-dn.net/?f=%5BFe%5E%7B%2B3%7D%5D%3D0.700%20M)
![[NO_{3}^{-}]=2.10 M](https://tex.z-dn.net/?f=%5BNO_%7B3%7D%5E%7B-%7D%5D%3D2.10%20M)
Explanation:
Here, a solution of Fe(NO₃)₃ is diluted, as the total volume of the solution has increased. The formula for dilution of the compound is mathematically expressed as:

Here, C and V are the concentration and volume respectively. The numbers at the subscript denote the initial and final values. The concentration of Fe(NO₃)₃ is 1.75 M. As ferric nitrate dissociates completely in water, the initial concentration of ferric is also 1.75 M.
Solving for [Fe],
![[Fe^{+3}]=\frac{C_{1}.V_{1}}{V_{2} }](https://tex.z-dn.net/?f=%5BFe%5E%7B%2B3%7D%5D%3D%5Cfrac%7BC_%7B1%7D.V_%7B1%7D%7D%7BV_%7B2%7D%20%7D)
![[Fe^{+3}]=\frac{(1.75).(30.0)}{45.0+30.0 }](https://tex.z-dn.net/?f=%5BFe%5E%7B%2B3%7D%5D%3D%5Cfrac%7B%281.75%29.%2830.0%29%7D%7B45.0%2B30.0%20%7D)
![[Fe^{+3}]=0.700 M](https://tex.z-dn.net/?f=%5BFe%5E%7B%2B3%7D%5D%3D0.700%20M)
For [NO₃⁻],
There are three moles of nitrate is 1 mole of Fe(NO₃)₃. This means that the initial concentration of nitrate ions will be three times the concentration of ferric nitrate i.e., it will be 5.25 M.
![[NO_{3}^{-}]=\frac{C_{1}.V_{1}}{V_{2} }](https://tex.z-dn.net/?f=%5BNO_%7B3%7D%5E%7B-%7D%5D%3D%5Cfrac%7BC_%7B1%7D.V_%7B1%7D%7D%7BV_%7B2%7D%20%7D)
![[NO_{3}^{-}]=\frac{(5.25)(30.0)}{30.0+45.0 }](https://tex.z-dn.net/?f=%5BNO_%7B3%7D%5E%7B-%7D%5D%3D%5Cfrac%7B%285.25%29%2830.0%29%7D%7B30.0%2B45.0%20%7D)
![[NO_{3}^{-}]=2.10 M](https://tex.z-dn.net/?f=%5BNO_%7B3%7D%5E%7B-%7D%5D%3D2.10%20M)
Answer:
B
Explanation:
because mass of an electron is 1/2000.
Answer:
proton
Explanation:
neutron: A subatomic particle forming part of the nucleus of an atom. It has no charge. It is equal in mass to a proton or it weighs 1 amu.
Answer:
2.70 M.
Explanation:
Molar mass of MgO = 24.3 + 16 = 40.3.
A molar solution will contain 40.3g in 1 Litre.
So 800 mL of 1 molar solution will contain 0.8* 40.3 g MgO.
So the molarity of the given solution
= 87 / (0.8*40.3)
= 2.699 M.