The standard formation equation for glucose C6H12O6(s) that corresponds to the standard enthalpy of formation or enthalpy change ΔH°f = -1273.3 kJ/mol is
C(s) + H2(g) + O2(g) → C6H12O6(s)
and the balanced chemical equation is
6C(s) + 6H2(g) + 3O2(g) → C6H12O6(s)
Using the equation for the standard enthalpy change of formation
ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)
ΔHoreaction = ΔHfo[C6H12O6(s)] - {ΔHfo[C(s, graphite) + ΔHfo[H2(g)] + ΔHfo[O2(g)]}
C(s), H2(g), and O2(g) each have a standard enthalpy of formation equal to 0 since they are in their most stable forms:
ΔHoreaction = [1*-1273.3] - [(6*0) + (6*0) + (3*0)]
= -1273.3 - (0 + 0 + 0)
= -1273.3
Answer:
42 liters of oxygen (liquid) weighs 47900 grams.
Explanation:
First, let's write the givens in the form of a chemical equation:
3A + B ...................> 4X + 2Y
Now we find that this equation implies the following:
For every 4X and 2Y formation, 3A and 1B must disappear (react).
Comparing this implication to the above choices, we find that the right answer is: <span>The rate of formation of X is four times the rate of disappearance of B.</span>
The mass of magnesium in 
atoms is 240 g.
Answer: Option A
<u>Explanation:</u>
First, we have to convert the atoms to moles of magnesium.
We know that atoms are present in 1 mole of magnesium. So,
Thus,
Thus, 240 g of Magnesium is present in atoms.
