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
![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,
![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 EMPIRICAL FORMULA OF THE SUBSTANCE IS C2H5NO
Explanation:
The steps involved in calculating the empirical formula of this substance in shown in the table below:
Element Carbon Hydrogen Nitrogen Oxygen
1. % Composition 40.66 8.53 23.72 27.09
2. Mole ratio =
%mass/ atomic mass 40.66/12 8.53/1 23.72/14 27.09/16
= 3.3883 8.53 1,6943 1.6931
3. Divide by smallest
value (0.6931) 3.3883/1.6931 8.53/1.6931 1.6943/1.6931 1.6931/1.6931
= 2.001 5.038 1.0007 1
4. Whole number ratio 2 5 1 1
The empirical formula = C2H5NO
Answer:
209.98 g of NaOH
Explanation:
We are given;
- Volume of HCl as 3 L
- Molarity of HCl as 1.75 M
We are required to calculate the mass of NaOH required to completely neutralize the acid given.
First, we write a balanced equation for the reaction between NaOH and HCl
That is;
NaOH + HCl → NaCl + H₂O
Second, we determine the number of moles of HCl
Number of moles = Molarity × Volume
= 1.75 M × 3 L
= 5.25 moles
Third, we use the mole ratio to determine the moles of NaOH
From the reaction,
1 mole of NaOH reacts with 1 mole of HCl
Therefore;
Moles of NaOH = Moles of HCl
= 5.25 moles
Fourth, we determine the mass of NaOH
Molar mass of NaOH = 39.997 g/mol
Mass of NaOH = 5.25 moles × 39.997 g/mol
= 209.98 g
Thus, 209.98 g of NaOH will completely neutralize 3L of 1.74 M HCl
They have the same amount but if you count the wrist it was more then the leg and ankle.
