Below the ecosystem is soil, granted, and bedrock. however for the ocean ecosystem has sand below it.
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
Answer: 568g/mol
Explanation:
It should be noted that there are 40 atoms of carbon in lycopene.
Since mass of 1 carbon = 12g/mol
Mass of 40 carbon atoms = 40 × 12g/mol = 480g/mol
Let the molar mass of lycopene be represented by x.
Therefore the molar mass of carbon = x × mass percent of carbon in lycopene
x × 84.49% = 480g/mol
x × 0.8449 = 480g/mol
x = 480/0.8449
x = 568g/mol
The molar mass of lycopene is 568g/mol
