Answer:- d. London dispersion
Explanations:- Carbon dioxide structure is given below:
O=C=O
It has two polar C=O bonds but even then it is non polar molecule as it's linear and the dipole moment of one C=O bond is canceled by it's opposite C=O bond.
being non polar, the only attraction forces present in it are London dispersion forces.
You can infer that the size of sugar particles that are dissolved in a mixture of sugar and water are very small since you cannot see the sugar particles. Sugar is an organic compound which is held by covalent bonding. When we dissolve sugar in water, it does not ionizes or dissociates rather it stays as a sugar molecule dissolved in the solvent which is water. These particles are very which is impossible to be seen by the naked eye. Dissolution involves the solute particles and solvent particles interacting and attracting each other forming a solution. Sugar molecules are attracted to water since it is a polar molecule.
Answer:
14.3mL you require to reach the half-equivalence point
Explanation:
A strong acid as HClO₄ reacts with a weak base as CH₃CH₂NH₂, thus:
CH₃CH₂NH₂ + HClO₄ → CH₃CH₂NH₃⁺ + ClO₄⁻
As the reaction is 1:1, to reach the equivalence point you require to add the moles of HClO₄ equal to moles CH₃CH₂NH₂ you add originally. Also, half-equivalence point requires to add half-moles of CH₃CH₂NH₂ you add originally.
Initial moles of CH₃CH₂NH₂ are:
20.8mL = 0.0208L × (0.51mol CH₃CH₂NH₂ / 1L) =
0.0106moles CH₃CH₂NH₂
To reach the half-equivalence point you require:
0.0106moles ÷ 2 = 0.005304 moles HClO₄
As concentration of HClO₄ is 0.37M, volume you require to add 0.005304moles is:
0.005304 moles HClO₄ ₓ (1L / 0.37mol) = 0.0143L =
<h3> 14.3mL you require to reach the half-equivalence point</h3>
Answer:
Between 5.5 to 7.5
The optimum pH range for most plants is between 5.5 and 7.5