Answer: HYDROGEN BONDS
Explanation:
Water molecules attract each other happily thanks to their polarity. A hydrogen atom plus end associates an oxygen atom minus end.
These attractions are an example of hydrogen bonds, weak interactions forming between a partially positive charged hydrogen and a more electronegative atom like oxygen. The hydrogen atoms involved in bonding with hydrogen need to be bound to electronegative atoms such as Oxygen and fluorine
Answer : The volume of solution will be 2.96 liters.
Explanation :
Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.
Formula used :

In this question, the solute is NaF.
Now put all the given values in this formula, we get:



Therefore, the volume of solution will be 2.96 liters.
Answer:
See explanation and image attached
Explanation:
This reaction is known as mercuric ion catalyzed hydration of alkynes.
The first step in the reaction is attack of the mercuric ion on the carbon-carbon triple bond, a bridged intermediate is formed. This bridged intermediate is attacked by water molecule to give an organomercury enol. This undergoes keto-enol tautomerism, proton transfer to the keto group yields an oxonium ion, loss of the mercuric ion now gives equilibrium keto and enol forms of the compound. The keto form is favoured over the enol form.
PH and conductivity have no common rekationship
Answer:
E° = 0.65 V
Explanation:
Let's consider the following reductions and their respective standard reduction potentials.
Sn⁴⁺(aq) + 2 e⁻ → Sn²⁺(aq) E°red = 0.15 V
Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
The reaction with the highest reduction potential will occur as a reduction while the other will occur as an oxidation. The corresponding half-reactions are:
Reduction (cathode): Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V
Oxidation (anode): Sn²⁺(aq) → Sn⁴⁺(aq) + 2 e⁻ E°red = 0.15 V
The overall cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = 0.80 V - 0.15 V = 0.65 V