The best and most correct answer among the choices provided by your question is the third choice or letter C.
Premenstrual syndrome<span> (</span>PMS<span>) has a wide variety of </span>symptoms, including mood swings, tender breasts, food cravings, fatigue, irritability and depression. It's estimated that as many as 3 of every 4 menstruating women have experienced some form of premenstrual syndrome<span>. </span>Symptoms<span> tend to recur in a predictable pattern.</span>
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On eof the forces is inertia. This can happen when you move your furniture to clean under them. When you move an object in a straight line without chanign its composition (e.g moving a chair or a table from one side to the other) you are working with inertia. The secon force is friction. It s present when you clean a surface by rubbing it with a tool like a damp cloth or a mop.
Glucose can be converted into ethanol, carbon dioxide and energy (usually in the form of heat).
Using chemical formulas, this can be illustrated as follows:
Glucose = Ethanol + Carbon Dioxide + Energy
C6H12O6 = 2C2H5OH + 2CO2 + Energy
From the periodic table:
molecular mass of carbon = 12 grams
molecular mass of oxygen = 16 grams
molecular mass of hydrogen = 1 gram
Therefore:
molar mass of glucose = 6(12) + 12(1) + 6(16) = 180 grams
molar mass of ethanol = 2(12) + 5(1) + 16 + 1 = 46 grams
molar mass of carbon dioxide = 1(12) + 2(16) = 44 grams
Based on the conversion equation, each one mole of glucose converts into two moles of ethanol and two moles of carbon dioxide.
Therefore, each 180 grams of glucose converts into 46 x 2 = 92 grams of ethanol and 44 x 2 = 88 grams of carbon dioxide in addition to energy.
To calculate the mass of ethanol and carbon dioxide produced from 200 grams of glucose, we will simply use cross multiplication as follows:
mass of ethanol = (200 x 92) / 180 = 102.2 grams
mass of carbon dioxide = (200 x 88) / 180 = 97.7 grams
Total mass of ethanol and carbon dioxide = 102.2 + 97.7 = 199.9 grams
Answer:
the standard enthalpy of formation of this isomer of C₈H₁₈ (g) = -375 kj/mol
Explanation:
The given combustion reaction
C₈H₁₈ + 25/2 O₂ → 8 CO₂ + 9 H₂O ............................(1)
Heat of reaction or enthalpy of combustion = -5099.5 kj/mol
from equation (1)
ΔH⁰reaction = (Enthalpy of formation of products - Enthalpy of formation of reactants)
Or, - 5099.5 = [8 x ΔH⁰f(CO₂) +9 x ΔH⁰f(H₂O)] - [ΔH⁰f(C₈H₁₈) + ΔH⁰f(O₂)].................................(2)
Given ΔH⁰f(CO₂) = - 393.5 kj/mol & ΔH⁰f(H₂O) = - 285.8 kj/mol and ΔH⁰f(O₂)= 0
Using equation (2)
ΔH⁰f(C₈H₁₈) = -621 kj/mol
