To get the molarity, you divide the moles of solute by the litres of solution.
Molarity
=
moles of solute
litres of solution
For example, a 0.25 mol/L NaOH solution contains 0.25 mol of sodium hydroxide in every litre of solution.
To calculate the molarity of a solution, you need to know the number of moles of solute and the total volume of the solution.
To calculate molarity:
Calculate the number of moles of solute present.
Calculate the number of litres of solution present.
Divide the number of moles of solute by the number of litres of solution.
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>
False. An increase in temperature is an exothermic reaction. However, when a temperature decreases this is known as an endothermic reactionz
They bond because they want to make their outer electron shells more stable
Hope this helps
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Unsaturation (IHD) 2 hydrogen Needed
IHD = [(2n+2) -H]/2
(H: X=1, N=-1, O= zero)
Unsaturation:
Double bonds = 1
Rings = 1
Triple Bonds = 2
The degrees of unsaturation in a molecule are additive — a
molecule with one double bond has one degree of unsaturation, a molecule with
two double bonds has two degrees of unsaturation, and so forth.
