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:
2,4,4-trimethyl-2-pentene yields mixture of 
and 
Explanation:
In ozonolysis (hydrolysis step involve a reducing agent such as Zn, 
etc.), a pi bond is broken to form ketone/aldehyde.
Ketone is formed from di-substituted side of double bond and aldehyde is formed from mono-substituted side of double bond.
Ozoznolysis involves two consecutive steps : (1) formation of ozonide, (2) hydrolysis of ozonide.
Hydrolysis can be done with/without using reducing agent. Carboxylic acid/carbon dioxide/ketone is produced when hydrolysis is done without using reducing agent.
Here, 2,4,4-trimethyl-2-pentene yields mixture of and
Reaction steps are shown below.
The classification of it being a metal, nonmetal, or metalliod will be useful in the process of elimination to determine what it is. Then for the second test, meauring the atomin radius will narrow it down quicker to the mystery elemet's name.
Since you determined what part of the periodic table it's on, then when measuring the atomic radius, you should be able to pinpoint what the element is more surely.
Answer: False
Explanation: just took on edge
Answer:
Explanation:
Hello,
Based on the given undergoing chemical reaction is is rewritten below:
By stoichiometry we find the minimum mass of H2SO4 (in g) as shown below:
Moreover, mass of H2 gas (in g) would be produced by the complete reaction of the aluminum block turns out:
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