Answer:
endothermic
Explanation:
condensate is an endothermic change
Answer:
0.004548 M is the concentration of B at equilibrium at 500 K.
Explanation:
A(aq) ⇆ 2 B(aq)
Initially 3.00 M
At equilibrium 3.00 -x 2x
Equilibrium constant of the reaction at 500 K =
Concentration of A at 500 K at equilibrium , [A] = (3.00 -x )M
Concentration of B at 500 K at equilibrium,[B]= 2x
An expression of equilibrium constant is given as:
![K_c=\frac{[B]^2}{[A]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BB%5D%5E2%7D%7B%5BA%5D%7D)
On solving for x:
x = 0.002274 M
[B] = 2 x = 2 × 0.002274 M = 0.004548 M
[A] = (3-x) = 3 M - 0.002274 M =2.997726 M
0.004548 M is the concentration of B at equilibrium.
Answer: Molecule
Step-by-step-explanation
Answer:
ΔH°(f) = -110.5 Kj/mole (exothermic)
Explanation:
C + 1/2O₂ => CO
This is asking for the 'Standard Heat of Formation (ΔH°(f)* for carbon monoxide (CO). Values for many compounds can be found in the appendix of most college general chemistry text books. From Ebbing & Gammon, 11th edition, General Chemistry, Appendix C, page 8A.
*Standard Heat of Formation by definition is the heat gained or lost on formation of a substance (compound) from its basic elements in standard state.
The ΔH°(f) values as indicated are found in the appendix of most college chemistry texts. By choosing any compound, one can determine the standard heat of formation equation for the substance of interest. For example, consider Magnesium Carbonate; MgCO₃(s).The basic standard states of each element is found in the Appendix on Thermodynamic Properties for Substances at 25°C & 1 atm. having ΔH°(f) values = 0.00 Kj/mole. All elements in standard state have a 0 Kj/mol. See appendix and note that under the ΔH°(f) symbol some substances have 0.00 Kj/mol values. The associated element will be in basic standard state,
Standard Heat of Formation Equation for formation of Magnesium Carbonate;
Mg°(s) + C°(gpt)* + 3/2O₂(g) => MgCO₃(s) ; ΔH°(f) = -1111.7 Kj/mole
* gpt => graphite
