To determine the mass of the sample, first find the volume difference after and before the aluminum was placed, the volume change is equal to the volume of the submerged object, in this case aluminum.
Then knowing volume of aluminum and the density of it, we can solve for the mass.
D = m/v
Dv = m
2.7 g/ml • 8 ml = 21.6 grams.
Aldehydes and ketones having
α-hydrogen atoms, undergoes aldol condensation, in present of base (NaOH).
The initial product formed during this reaction is
β-hydroxy alcohol, which then undergoes dehydration to form
α,β-unsaturated aldehyde or ketone.
In present case, 3,3-dimethyl-2-butanone has 2α-hyrogen atom, while methylcyclopentane-1-carbaldehyde has 1α-hydrogen atom. So the major product formed during cross aldol condensation reaction of these reactants is:
5-hydroxy-4,4-dimethly-1-(2-methylcyclopentyl)pent-1-en-3-one. The complete reaction product formed is shown below.
Answer:
a. 5.36x10⁻⁴ g/mL
b. 4.29x10⁻⁵ g/mL
Explanation:
As the units for concentration are not specified, I'll respond using g/mL.
a. We <em>divide the sample mass by the final volume</em> in order to <u>calculate the concentration</u>:
- 0.268 g / 500 mL = 5.36x10⁻⁴ g/mL
b. We can use C₁V₁=C₂V₂ for this question:
- 8.00 mL * 5.36x10⁻⁴ g/mL = C₂ * 100.00 mL
Answer:
Explanation: Does any of this look right? To be honest I just looked it up.
Where are the following answers?