Elements in the third row can break the octet rule
Transpiration is the progression of <em>water </em>inside a plant! So, the molecule representing transpiration is going to be good ol' H2O! =)
Answer:
Here's what I get.
Explanation:
At the end of the reaction you will have a solution of the alcohol in THF.
The microdistillation procedure will vary, depending on the specific apparatus you are using, but here is a typical procedure.
- Transfer the solution to a conical vial.
- Add a boiling stone.
- Attach a Hickman head (shown below) and condenser.
- Place the assembly in in the appropriate hole of an aluminium block on top of a hotplate stirrer.
- Begin stirring and heating at a low level so the THF (bp 63 °C) can distill slowly.
- Use a Pasteur pipet to withdraw the THF as needed.
- When all the THF has been removed, raise the temperature of the Al block and distill the alcohol (bp 143 °C).
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.
