The acid clouds fog the important nutrients like leave and needles . this loss of nutrient make it easier for infections , insects , and cold weather to damage trees and forests
Answer:
A solution in which no more solute can be dissolved in is referred to as SATURATED. In such a solution, the concentration of solute is called SOLUBILITY . When that concentration is reported in moles per liter, it is more specifically called MOLAR SOLUBILITY. A special equilibrium constant called the SOLUBILITY PRODUCT constant is calculated from the molar concentrations of the aqueous components of the dissolution equation.
Explanation:
The solubility of a solute in a solvent is the maximum amount of solute in moles that will be dissolved in 1dm3 of the solvent at a specified temperature. Once the maximum number or concentration has been reached, the solvent can no longer take in solutes and this point in the reaction, the solution is said to be saturated. That is the composition of the saturated solution is not affected by the presence of excess solute. An unsaturated solution has a lower concentration of solute and can dissolve more solutes if added until it becomes saturated.
Solubility when reported in moles per liter is called molar solubility of the solution and it gives a more accurate measurement of yh solubility of a solution. The solubility product constant is calculated from the molar concentrations of the aqueous components of the dissolution equation. This solubility product constant explains the balance between dissolved ions from the salt and undissolved salt in a dissolution equation.
The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ of ) =-1275.0
enthalpy of combustion of oxygen(Δ of ) = zero
enthalpy of combustion of carbon dioxide(Δ of ) = -393.5
enthalpy of combustion of water(Δ of ) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ (products) - ∈Δ (reactants)
(s) +6 (g) → 6 (g)+ 6 (l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
If you have a book, read it!! I promise you, it tells you this answer!
Explanation:
As the charge of all electrons are equal, the repulsive force exerted by each of them is also going to be equal. So, as K has more electrons repulsing its valence electron than Na, it has greater electron shielding.