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>
C
explanation; ( i’m smart )
Answer:
= 97.44 Liters at S.T.P
Explanation:
The reaction between Iron (iii) oxide and Carbon monoxide is given by the equation;
Fe2O3(s)+ 3CO(g) → 3CO2(g) + 2Fe(s)
From the reaction when the reactants react, 2 moles of Fe and 3 moles of CO2 are produced.
Therefore; Mole ratio of Iron : Carbon dioxide is 2:3
Thus; Moles of Carbon dioxide = (2.9/2)×3
= 4.35 moles
But; 1 mole of CO2 at s.t.p occupies 22.4 liters
Therefore;
Mass of CO2 = 22.4 × 4.35 Moles
= 97.44 L
Answer:
295.7 mL of 24% trichloroacetic acid (tca) is needed .
Explanation:
Let the volume of 24% trichloroacetic acid solution be x
Volume of required 10% trichloroacetic acid solution =8 bottles of 3 ounces
= 24 ounces = 709.68 mL
(1 ounces = 29.57 mL)
Amount of trichloroacetic acid in 24% solution of x volume of solution will be equal to amount of trichloroacetic acid in 10% solution of volume 709.68 mL.
x = 295.7 mL
295.7 mL of 24% trichloroacetic acid (tca) is needed .
The amount or disorder in a chemical system or system is measured by it's Entropy. Entropy is the simply the measure of the amount of disorder or lack of structure present within a system.
