Addition of water to an alkyne gives a keto‑enol tautomer product and that is the product changed into 2-pentanone, then the alkyne need to had been 1-pentyne. 2-pentyne might have given a combination of 2- and 3-pentanone.
<h3>
What is the keto-enol means in tautomer?</h3>
They carries a carbonyl bond even as enol implies the presence of a double bond and a hydroxyl group. The keto-enol tautomerization equilibrium is depending on stabilization elements of each the keto tautomer and the enol tautomer.
- The enol that could provide 2-pentanone might had been pent-1- en - 2 -ol. Because an equilibrium favors the ketone so greatly, equilibrium isn't an excellent description.
- If the ketone have been handled with bromine, little response might be visible because the enol content material might be too low.
- If a catalyst have been delivered, NaOH for example, then formation of the enolate of pent-1-en - 2 - ol might shape and react with bromine.
- This might finally provide a bromoform product. Under acidic conditions, the enol might desire formation of the greater substituted enol constant with alkene stability.
Answer:
m = 0.531 molal
Explanation:
∴ m fructose = 3.35 g
∴ V water = 35.0 mL
∴ ρ H2O = 1 g/mL
- molality = moles solute / Kg solvent
∴ Mw fructose = 180.16 g/mol
⇒ moles fructose = 3.35 g * ( mol / 180.16 g) = 0.0186 mol fructose
⇒ m H2O = 35.0 mL * ( 1 g/mL ) * ( Kg/1000g) = 0.035 Kg H2O
⇒ molality (m) = 0.0186 mol fructose / 0.035 Kg H2O
⇒ m = 0.531 molal
Answer:
Δ[NH₃]/Δt = 2/3 ( Δ[H₂]/Δt )
Explanation:
For determining rates as a function of reaction coefficients one should realize that these type problems are <u>always in pairs</u> of reaction components. For the reaction N₂ + 3H₂ => 2NH₃ one can compare ...
Δ[N₂]/Δt ∝ Δ[H₂]/Δt, or
Δ[N₂]/Δt ∝ Δ[NH₃]/Δt, or
Δ[H₂]/Δt ∝ Δ[NH₃]/Δt, but never 3 at a time.
So, set up the relationship of interest ( ammonia rate vs. hydrogen rate)... nitrogen rate is ignored.
Δ[H₂]/Δt ∝ Δ[NH₃]/Δt
Now, 'swap' coefficients of balanced equation and apply to terms given then set term multiples equal ...
N₂ + 3H₂ => 2NH₃ => 2(Δ[H₂]/Δt) = 3(Δ[NH₃]/Δt) => 2/3(Δ[H₂]/Δt) = (Δ[NH₃]/Δt)
NOTE => Comparing rates individually of the component rates in reaction process, the rate of H₂(g) consumption is 3/2 times <u>faster</u> than NH₃(g) production (larger coefficient). So, in order to compose an equivalent mathematical relationship between the two, one must reduce the rate of the H₂(g) by 2/3 in order to equal the rate of NH₃(g) production. Now, given the rate of one of the components as a given, substitute and solve for the unknown.
CAUTION => When Interpreting rate of reaction one should note that the rate expression for an individual reaction component defines 'instantaneous' rate or speed. <u>This means velocity (or, speed) does not have 'signage'</u>. Yes, one may say the rate is higher or lower as time changes but that change is an acceleration or deceleration for one instantaneous velocity to another. Acceleration and Deceleration do have signage but the positional instantaneous velocity (defined by a point in time) does not. Thus is reason for the 'e-choice' answer selection without the signage associated with the expression terms.
Answer:
See explanation
Explanation:
All physical quantities are broadly classified into scalar quantities and vector quantities.
Scalar quantities have magnitude but do not have direction. Vector quantities have both magnitude and direction. Hence the main difference between a scalar quantity and a vector quantity is that a vector has direction while a scalar quantity does not.
Speed is a scalar quantity while velocity is a vector quantity. You have just specified a magnitude of 200 mph without mentioning its direction. This implies that you are referring to speed of the cars in the race and not velocity because velocity of the cars must indicate the direction!
The molecular formula of a compound that has 6 carbon, 8 hydrogen, and 2 oxygen will be c6h8o2
