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>
Endothermic - takes in heat from the surroundings
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Answer: There are 0.0637 moles present in 85.0 mL of 0.750 M KOH.
Explanation:
Given: Volume = 85.0 mL (1 mL = 0.001 L) = 0.085 L
Molarity = 0.750 M
It is known that molarity is the number of moles of solute present in liter of a solution.
Therefore, moles present in given solution are calculated as follows.
Thus, we can conclude that there are 0.0637 moles present in 85.0 mL of 0.750 M KOH.
That's false. A melting point is hot/warm, while the freezing point is cold.
