Is cloud seeding a physical or chemical process?

AlF3 is almost insoluble in anhydrous HF but dissolves if KF is present.passage of BF3 through the resulting solution causes AlF

miskamm [114]

It's because the product formed with BF3 is more complex which able to decompose AlF3.

AlF3 doesn't dissolve in HF because of the fluorine. It's doesn't allow for coordination due to the hydrogen. However, it will dissolve in KF. If you look at the chemical reaction, it's able to form a salt.

3KF+AlF3−>3KF.AlF3

4 0

4 years ago

What enzyme catalyses the reaction of PRPP transfer to hypoxanthine with formation of IMP?

Alona [7]

Answer:

HPRT

Explanation:

HPRT catalyzes the salvage reactions of hypoxanthine and guanine with PRPP to form IMP and GMP

The formation of GMP from IMP requires oxidation at C-2 of the purine ring, followed by a glutamine-dependent amidotransferase reaction that replaces the oxygen on C-2 with an amino group to yield 2-amino,6-oxy purine nucleoside monophosphate, or as this compound is commonly known, guanosine monophosphate.

8 0

3 years ago

The emission of the light waves in the lasers allows them to overlap properly to create larger waves, which means that _____ ove

Vikentia [17]

The emission of the light waves ..................... which means that THE TROUGHS OF ONE WAVE AND THE CREST OF ANOTHER overlap.
When the trough and the crest of wave overlap, it results in constructive interference and the amplitude of the two waves are added together to form a larger amplitude.

5 0

4 years ago

2) Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H2

nadezda [96]

Answer:

Kc = 3.90

Explanation:

CO reacts with $H_2$ to form $CH_4$ and $H_2O$ . balanced reaction is:

$CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)$

No. of moles of CO = 0.800 mol

No. of moles of $H_2$ = 2.40 mol

Volume = 8.00 L

Concentration = $\frac{Moles}{Volume\ in\ L}$

Concentration of CO = $\frac{0.800}{8.00} = 0.100\ mol/L$

Concentration of $H_2$ = $\frac{2.40}{8.00} = 0.300\ mol/L$

$CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g) + H_2O(g)$

Initial 0.100 0.300 0 0

equi. 0.100 -x 0.300 - 3x x x

It is given that,

at equilibrium $H_2O (x)$ = 0.309/8.00 = 0.0386 M

So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M

At equilibrium $H_2$ = 0.300 - 0.0386 × 3 = 0.184 M

At equilibrium $CH_4$ = 0.0386 M

$Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}$

$Kc=\frac{0.0386 \times 0.0386}{(0.184)^3 \times 0.0614} =3.90$

8 0

4 years ago

Write a mechanism for the esterification of propanoic acid with 18O-labeled ethanol. Show clearly the fate of the 18O label. (b)

tatuchka [14]

Answer:

See explanation and images attached

Explanation:

a) In the mechanism for the acid catalysed esterification of propanoic acid using ethanol, we can see that the first step is the protonation of the acid followed by nucleophillic attack of the alcohol. Loss of water and consequent deprotonation regenerates the acid catalyst. We can see the fate of the 18O labelled ethanol in the mechanism shown.

b) In the second mechanism, an unnamed ester is hydrolysed using an acid catalyst. The attack of the acid and subsequent nucleophillic attack of water labelled with 18O leads to the incorporation of this 18O into the product acid as shown in the mechanism attached to this answer.

5 0

3 years ago