Answer: acetone is a ketone and not an aldehyde. Therefore it's false
Explanation:
Acetone is of structure CH3-CO-CH3
which makes it a ketone.
Both aldehyde and ketone have a carbonyl group but there is difference between aldehyde and ketone.
Aldehydes have a general formula R-CHO
where R is a methyl group including H atom, and CHO is the functional group.
While
Ketone have the general formula R-CO-Ri
where R and Ri are methyl groups excluding H atom, and CO is the functional group.
Hint: ketones usually end with the suffix "one" as in acetone.
Species that have a lone pair of electrons often donate electrons by resonance while substituents that are electron deficient take away electrons by resonance.
<h3>What is resonance?</h3>
The term resonace has to do with the movement of electron pairs in a molecule. Inductive effects has to do with the drawing of electron density towards an atom or bond.
The two effects depends on the nature of a substituent. For instance, species that have a lone pair of electrons often donate electrons by resonance while substituents that are electron deficient take away electrons by resonance.
The question is incomplete hence the exact nature of the substituents can not be determined.
Learn more about resonance: brainly.com/question/23287285?
I think that building on a green hill
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:
25%
Explanation:
Half life means that 50% of the sample is gone at 3.3 years. This means that an additional 3.3 years (total 6.6 years) will reduce the sample a further 50% from the point at 3.3 years. In numbers, this means 50% of 50% (0.50*0.50), which is 25%.
