Answer:
6.25 μg/mL
Explanation:
When a dilution is made, the mass of the solute is conserved (Lavoiser's law), so the mass pipetted will be the mass in the assay. The mass is the concentration (C) multiplied by the volume (V). If the pipet solution is called 1, and the assay 2:
m1 = m2
C1*V1 = C2*V2
C1 = 250 μg/mL
V1 = 25 μL
V2 = 975 μL + 25 μL = 1000 μL (is the final volume of the assay after the addition of LDH)
250*25 = C2*1000
C2 = 6.25 μg/mL
Answer:
![[Ag^{+}]=4.2\times 10^{-2}M](https://tex.z-dn.net/?f=%5BAg%5E%7B%2B%7D%5D%3D4.2%5Ctimes%2010%5E%7B-2%7DM)
Explanation:
Given:
[AgNO3] = 0.20 M
Ba(NO3)2 = 0.20 M
[K2CrO4] = 0.10 M
Ksp of Ag2CrO4 = 1.1 x 10^-12
Ksp of BaCrO4 = 1.1 x 10^-10
![BaCrO_4 (s)\leftrightharpoons Ba^{2+}(aq)\;+\;CrO_{4}^{2-}(aq)](https://tex.z-dn.net/?f=BaCrO_4%20%28s%29%5Cleftrightharpoons%20%20Ba%5E%7B2%2B%7D%28aq%29%5C%3B%2B%5C%3BCrO_%7B4%7D%5E%7B2-%7D%28aq%29)
![Ksp=[Ba^{2+}][CrO_{4}^{2-}]](https://tex.z-dn.net/?f=Ksp%3D%5BBa%5E%7B2%2B%7D%5D%5BCrO_%7B4%7D%5E%7B2-%7D%5D)
![1.2\times 10^{-10}=(0.20)[CrO_{4}^{2-}]](https://tex.z-dn.net/?f=1.2%5Ctimes%2010%5E%7B-10%7D%3D%280.20%29%5BCrO_%7B4%7D%5E%7B2-%7D%5D)
![[CrO_{4}^{2-}]=\frac{1.2\times 10^{-10}}{(0.20)}= 6.0\times 10^{-10}](https://tex.z-dn.net/?f=%5BCrO_%7B4%7D%5E%7B2-%7D%5D%3D%5Cfrac%7B1.2%5Ctimes%2010%5E%7B-10%7D%7D%7B%280.20%29%7D%3D%206.0%5Ctimes%2010%5E%7B-10%7D)
Now,
![Ag_{2}CrO_4(s) \leftrightharpoons 2Ag^{+}(aq)\;+\;CrO_{4}^{2-}(aq)](https://tex.z-dn.net/?f=Ag_%7B2%7DCrO_4%28s%29%20%5Cleftrightharpoons%20%202Ag%5E%7B%2B%7D%28aq%29%5C%3B%2B%5C%3BCrO_%7B4%7D%5E%7B2-%7D%28aq%29)
![Ksp=[Ag^{+}]^{2}[CrO_{4}^{2-}]](https://tex.z-dn.net/?f=Ksp%3D%5BAg%5E%7B%2B%7D%5D%5E%7B2%7D%5BCrO_%7B4%7D%5E%7B2-%7D%5D)
![1.1\times 10^{-12}=[Ag^{+}]^{2}](6.0\times 10^{-10})](https://tex.z-dn.net/?f=1.1%5Ctimes%2010%5E%7B-12%7D%3D%5BAg%5E%7B%2B%7D%5D%5E%7B2%7D%5D%286.0%5Ctimes%2010%5E%7B-10%7D%29)
![[Ag^{+}]^{2}]=\frac{1.1\times 10^{-12}}{(6.0\times 10^{-10})}= 1.8\times 10^{-3}](https://tex.z-dn.net/?f=%5BAg%5E%7B%2B%7D%5D%5E%7B2%7D%5D%3D%5Cfrac%7B1.1%5Ctimes%2010%5E%7B-12%7D%7D%7B%286.0%5Ctimes%2010%5E%7B-10%7D%29%7D%3D%201.8%5Ctimes%2010%5E%7B-3%7D)
![[Ag^{+}]=\sqrt{1.8\times 10^{-3}}=4.2\times 10^{-2}M](https://tex.z-dn.net/?f=%5BAg%5E%7B%2B%7D%5D%3D%5Csqrt%7B1.8%5Ctimes%2010%5E%7B-3%7D%7D%3D4.2%5Ctimes%2010%5E%7B-2%7DM)
So, BaCrO4 will start precipitating when [Ag+] is 4.2 x 1.2^-2 M
Answer:
The molecular formula of the compound is
. The molecular formula is obtained by the following expression shown below
![\textrm{Molecular formula }= n\times \textrm{Empirical formula}](https://tex.z-dn.net/?f=%5Ctextrm%7BMolecular%20formula%20%7D%3D%20n%5Ctimes%20%5Ctextrm%7BEmpirical%20formula%7D)
Explanation:
Given molecular mass of the compound is 176 g/mol
Given empirical formula is
Atomic mass of carbon, hydrogen and oxygen are 12 u , 1 u and 16 u respectively.
Empirical formula mass of the compound = ![\left ( 2\times12+4+16 \right ) \textrm{ u} = 44 \textrm{ g/mol}](https://tex.z-dn.net/?f=%5Cleft%20%28%202%5Ctimes12%2B4%2B16%20%5Cright%20%29%20%5Ctextrm%7B%20u%7D%20%3D%2044%20%5Ctextrm%7B%20g%2Fmol%7D)
![n = \displaystyle \frac{\textrm{Molecular formula mass}}{\textrm{Empirical formula mass}} \\n = \displaystyle \frac{176}{44} = 4](https://tex.z-dn.net/?f=n%20%3D%20%5Cdisplaystyle%20%5Cfrac%7B%5Ctextrm%7BMolecular%20formula%20mass%7D%7D%7B%5Ctextrm%7BEmpirical%20formula%20mass%7D%7D%20%5C%5Cn%20%3D%20%5Cdisplaystyle%20%5Cfrac%7B176%7D%7B44%7D%20%3D%204)
![\textrm{Molecular formula }= n\times \textrm{Empirical formula}](https://tex.z-dn.net/?f=%5Ctextrm%7BMolecular%20formula%20%7D%3D%20n%5Ctimes%20%5Ctextrm%7BEmpirical%20formula%7D)
Molecular formula = 4
Molecular formula is ![C_{8}H_{16}O_{4}](https://tex.z-dn.net/?f=C_%7B8%7DH_%7B16%7DO_%7B4%7D)
Answer:
The answer is evaporation
Answer:
pioneer species. because they are flower like species