Answer:
Concentration of nitrate in the new solution = 0.007 M
Explanation:
Given:
Concentration nitrate solution = 0.070 m
Volume of aliquote of the nitrate solution is add = 10.0 ml
Total volume = 100 ml
Find:
Concentration of nitrate in the new solution
Computation:
Number of M. mole = 0.070 m x 10.0 ml
Number of M. mole = 0.7 m-moles
Concentration of nitrate in the new solution = 0.7 m-moles / 100 ml
Concentration of nitrate in the new solution = 0.007 M
You have not mention here about those elements but the general concept for this is the:
uneven distribution ------> polar
even distribution -------> non-polar
Answer:
2.288 Moles of NH₄NO₃
Explanation:
The Balance chemical equation is as follow:
(NH₄)₂CO₃ + 2 HNO₃ → 2 NH₄NO₃ + H₂O + CO₂
To solve this problem we will do following steps:
Finding moles of Ammonium Carbonate:
As we know,
Moles = Mass / M.Mass
So,
Moles = 110 g / 96.08 g/mol
Moles = 1.144 moles
Calculating moles of Ammonium Nitrate:
According to balance chemical equation;
1 mole of (NH₄)₂CO₃ produces = 2 moles of NH₄NO₃
So,
1.144 moles of (NH₄)₂CO₃ will produce = X moles of NH₄NO₃
Solving for X,
X = 2 moles × 1.144 moles ÷ 1 mole
X = 2.288 moles of NH₄NO₃
C.
Protons is 8 so the atomic number will be 8.
Number of neutrons= atomic mass - atomic number
16-8 = 8.
Answer:
The rate at which 
is being produced is 0.0228 M/s.
The rate at which 
is being consumed is 0.0912 M/s.
Explanation:
Rate of the reaction : R
![R=\frac{-1}{4}\frac{d[PH_3]}{dt}=\frac{1}{6}\frac{d[H_2]}{dt}=\frac{1}{1}\frac{d[P_4]}{dt}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7B-1%7D%7B4%7D%5Cfrac%7Bd%5BPH_3%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B6%7D%5Cfrac%7Bd%5BH_2%5D%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B1%7D%5Cfrac%7Bd%5BP_4%5D%7D%7Bdt%7D)
The rate at which hydrogen is being formed = ![\frac{d[H_2]}{dt}=0.137 M/s](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5BH_2%5D%7D%7Bdt%7D%3D0.137%20M%2Fs)
![R=\frac{1}{6}\frac{d[H_2]}{dt}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7B1%7D%7B6%7D%5Cfrac%7Bd%5BH_2%5D%7D%7Bdt%7D)
The rate at which is being produced:
![R=\frac{1}{1}\frac{d[P_4]}{dt}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7B1%7D%7B1%7D%5Cfrac%7Bd%5BP_4%5D%7D%7Bdt%7D)
![0.0228 M/s=\frac{1}{1}\frac{d[P_4]}{dt}](https://tex.z-dn.net/?f=0.0228%20M%2Fs%3D%5Cfrac%7B1%7D%7B1%7D%5Cfrac%7Bd%5BP_4%5D%7D%7Bdt%7D)
The rate at which is being consumed :
![R=\frac{-1}{4}\frac{d[PH_3]}{dt}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7B-1%7D%7B4%7D%5Cfrac%7Bd%5BPH_3%5D%7D%7Bdt%7D)
![0.0228 M/s\times 4=\frac{-1}{1}\frac{d[PH_3]}{dt}](https://tex.z-dn.net/?f=0.0228%20M%2Fs%5Ctimes%204%3D%5Cfrac%7B-1%7D%7B1%7D%5Cfrac%7Bd%5BPH_3%5D%7D%7Bdt%7D)
![\frac{-1}{1}\frac{d[PH_3]}{dt}=0.912 M/s](https://tex.z-dn.net/?f=%5Cfrac%7B-1%7D%7B1%7D%5Cfrac%7Bd%5BPH_3%5D%7D%7Bdt%7D%3D0.912%20M%2Fs)
