Answer:
= 20.82 g of BaCl2
Explanation:
Given,
Volume = 200 mL
Molarity = 0.500 M
Therefore;
Moles = molarity × volume
= 0.2 L × 0.5 M
= 0.1 mole
But; molar mass of BaCl2 is 208.236 g/mole
Therefore; 0.1 mole of BaCl2 will be equivalent to;
= 208.236 g/mol x 0.1 mol
= 20.82 g
Therefore, the mass of BaCl2 in grams required is 20.82 g
Answer:
Relative volume of ether to water that should be used for the extraction = 1.205
Explanation:
The extraction/distribution coefficient of an arbitrary solvent to water for a given substance is expressed as the mass concentration of the substance in the arbitrary solvent (C₁) divided by the mass concentration of the substance in water (C₂).
K = (C₁/C₂)
Let the initial mass of the organic substance X in water be 1 g (it could be any mass basically, it is just to select a right basis, since we are basically working with percentages here).
If 94% of the organic substance X is extracted by ether in a single extraction, 0.94 g ends up in ether and 0.06 g of the organic substance X that remains in water.
Let the volume of ether required be x mL.
Let the volume of water required be y mL.
Relative volume of ether to water that should be used for the extraction = (x/y)
Mass concentration of the organic substance X in ether = (0.94/x)
Mass concentration of organic substance X in water = (0.06/y)
The distribution coefficient , Ko (Cether / C water), for an organic substance X at room temperature is 13.
13 = (0.94/x) ÷ (0.06/y)
13 = (0.94/x) × (y/0.06)
13 = (15.667y/x)
(x/y) = (15.667/13) = 1.205
Hope this Helps!!!
<span>3Ca + 2AlPO4 -> 2Al + Ca3(PO4)2 . . .the coefficient for Ca is 3.</span>
Answer:
Concentration of hydrogen ion, ![[H^+]=5.0118*10^{-6} M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D5.0118%2A10%5E%7B-6%7D%20M)
Explanation:
pH is defined as the negative logarithm of hydrogen ion's concentration.
The lower the value of pH, the higher the acidic the solution is.
The formula for pH can be written as:
![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
Given,
pH of the saliva of Marco = 5.3
To calculate: Hydrogen ion concentration in the saliva
Thus, applying in the formula as:
![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
![5.3=-log[H^+]](https://tex.z-dn.net/?f=5.3%3D-log%5BH%5E%2B%5D)
So,
![log[H^+]=-5.3](https://tex.z-dn.net/?f=log%5BH%5E%2B%5D%3D-5.3)
![[H^+]=10^{(-5.3)}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B%28-5.3%29%7D)
![[H^+]=5.0118*10^{-6} M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D5.0118%2A10%5E%7B-6%7D%20M)