The answer is "night sky"
hope i helped :)
<span>Colbat (ii) which is a compound birth out of the combination of chlorine and colbat to form Cocl2.6h2o has water in it as we can see from it's chemical it's hexahydrate
Anhydrous cobalt chloride as the word anhydrous clearly states , does not have water in</span>
Explanation:
Properties of a solution which are dependent on the ratio of number of solute particles to the number of solvent molecules in a solution are known as colligative properties.
Lowering of vapor pressure and elevation in boiling point are basically two of the colligative properties which indirectly help to measure the molecular weight of a substance.
Thus, we can conclude that molar mass of the solute can be determined by measuring lowering of vapor pressure and elevation in boiling point.
Answer : The ratio of the protonated to the deprotonated form of the acid is, 100
Explanation : Given,

pH = 6.0
To calculate the ratio of the protonated to the deprotonated form of the acid we are using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[Deprotonated]}{[Protonated]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D)
Now put all the given values in this expression, we get:
![6.0=8.0+\log \frac{[Deprotonated]}{[Protonated]}](https://tex.z-dn.net/?f=6.0%3D8.0%2B%5Clog%20%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D)
As per question, the ratio of the protonated to the deprotonated form of the acid will be:
Therefore, the ratio of the protonated to the deprotonated form of the acid is, 100
Answer:
Molarity of solution is 1.10x10⁻³ M
Explanation:
Solute NaOCl
7.4% by mass means, that in 100 grams of solution, we have 7.4 g of solute.
Molar mass of NaOCl = 74.45 g/m
Mol = Mass / Molar mass
7.4 g / 74.45 g/m = 0.099 moles
Density of solution = 1.12 g/mL
Density = Mass / volume
1.12g/mL = 100 g / volume
Volume = 100 g / 1.12g/mL = 89.3 mL
Molarity = mol /L
89.3 mL = 0.0893 L
0.099 moles / 0.0893 L = 1.10x10⁻³ M