1) is sulfuric acid
2)is nitrous acid
3)is hydrochlorous acid
4)is hydrobromous acid
5)is hydrophosphoric acid
6)is fluoric acid
7)is sulfuric acid
8)is chlorous acid
9)is nitric acid
10)is iodic acid
11)is acetous acid
12)is chlorous acid.
Based on the solubility observations, barium & aluminum could be distinguished by the addition of sodium chloride to the solutions.
<h3>What happens when NaCl is added to a solution?</h3>
- The ionic link that held sodium and chloride ions together is broken when water molecules force the ions apart.
- The sodium and chloride atoms are encircled by water molecules after the salt compounds are separated. After that, the salt dissolves and forms a homogenous solution.
- In order to keep patients from dehydrating, sodium chloride, an important nutrient, is employed in healthcare. It is employed as a spice to improve flavor and as a food preservative. Additionally, sodium chloride is employed in the production of polymers and other goods. Additionally, it is used to de-ice sidewalks and roadways.
- Adding water to sodium chloride results in a physical change because no new product is created.
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Hot-air balloons float because the air caught inside the balloon is heated up by a burner, making it less dense than the air outside. As the burner heats the air, it expands and some of the air escapes; that's what makes it less dense
Answer:
See explanation
Explanation:
The question is incomplete because the image of the alcohol is missing. However, I will try give you a general picture of the reaction known as hydroboration of alkenes.
This reaction occurs in two steps. In the first step, -BH2 and H add to the same face of the double bond (syn addition).
In the second step, alkaline hydrogen peroxide is added and the alcohol is formed.
Note that the BH2 and H adds to the two atoms of the double bond. The final product of the reaction appears as if water was added to the original alkene following an anti-Markovnikov mechanism.
Steric hindrance is known to play a major role in this reaction as good yield of the anti-Markovnikov like product is obtained with alkenes having one of the carbon atoms of the double bond significantly hindered.
Given buffer:
potassium hydrogen tartrate/dipotassium tartrate (KHC4H4O6/K2C4H4O6 )
[KHC4H4O6] = 0.0451 M
[K2C4H4O6] = 0.028 M
Ka1 = 9.2 *10^-4
Ka2 = 4.31*10^-5
Based on Henderson-Hasselbalch equation;
pH = pKa + log [conjugate base]/[acid]
where pka = -logKa
In this case we will use the ka corresponding to the deprotonation of the second proton i.e. ka2
pH = -log Ka2 + log [K2C4H4O6]/[KHC4H4O6]
= -log (4.31*10^-5) + log [0.0451]/[0.028]
pH = 4.15
