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Westkost [7]
3 years ago
10

gBe sure to answer all parts. Nicotine can be made when the following ammonium salt is treated with Na2CO3. Determine a stepwise

mechanism for this reaction. h5sm775 nicotine Part 1 out of 3 Choose the first step of the mechanism: h5sm775003 h5sm775002
Chemistry
1 answer:
xenn [34]3 years ago
6 0

Answer:

Kindly check the attached pictures.

Explanation:

For the reaction between ammonium salt and Na2CO3, in the reaction mechanism, there are going to be four steps for the production of nicotine. The chemical species to monitor here are the Na2CO3 and the starting material which has the amine group attached to it.

STEP ONE: Dissociation of the Na2CO3.

Na2CO3 <=========> 2Na^+ + CO3^2-

STEP TWO: (will be shown in the attached picture). It is the abstraction of H^+ from the amine in the starting material.

STEP THREE: (will be shown in the attached picture). It is the formation of a cyclic compound due to nucleophilic reaction.

STEP FOUR: (will be shown in the attached picture). It involve the the deprotonation and this is the final phase.

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What is the radius of a hydrogen atom whose electron is bound by 0.544 ev? express your answer with the appropriate units?
insens350 [35]
First, we need to calculate the principal quantum number n for this electron, using the equation:
E = (-13.60 eV) / (n x n)
where E is the energy that is used to bound the electron (here, E = - 0.544 eV).
- 0.544 eV = (-13.60 eV) / (n x n)
n x n = (- 13.60 eV) / (- 0.544 eV)
n x n = 25
n = 5

The orbital radius that is equal to the radius of a hydrogen atom is calculated using the equation:
r = 0.053 nm x n x n
r = 0.053 nm x 5 x 5
r = 0.053 nm x 25
r = 1.325 nm
6 0
3 years ago
9. At equilibrium a 2 L vessel contains 0.360
klio [65]

Answer:

Ke = 34570.707

Explanation:

  • H2(g) + Br2(g) → 2 HBr(g)

equilibrium constant (Ke):

⇒ Ke = [HBr]² / [Br2] [H2]

∴ [HBr] = (37.0 mol) / (2 L) = 18.5 mol/L

∴ [Br2] = (0.110 mol) / (2 L) = 0.055 mol/L

∴ [H2] = (0.360 mol) / (2 L) = 0.18 mol/L

⇒ Ke = (18.5 mol/L)² / (0.055 mol/L)(0.18 mol/L)

⇒ Ke = 34570.707

3 0
3 years ago
How many atoms of iodine are in 12.75g of CaI2? Hint. How many Iodines are there in one CaI2 particle?
xenn [34]

Answer:

5.225x10^{22}atoms\ I

Explanation:

Hello!

In this case, since 12.75 g of calcium iodide has the following number of moles (molar mass = 293.89 g/mol):

n_{CaI_2}=12.75gCaI_2*\frac{1molCaI_2}{293.89gCaI_2}=0.0434molCaI_2

In such a way, since 1 mole of calcium iodide contains 2 moles of atoms of iodine, and one mole of atoms of iodine contains 6.022x10²³ atoms (Avogadro's number), we compute the resulting atoms as shown below:

atoms\ I=0.0434molCaI_2*\frac{2molI}{1molCaI_2} *\frac{6.022x10^{23}atoms\ I}{1molI} \\\\atoms\ I = 5.225x10^{22}atoms\ I

Best regards!

4 0
3 years ago
Substance X has a fixed volume, and the attraction between its particles is strong. Substance Y has widely spread out particles
DedPeter [7]

Answer: X is a Solid; Y is a Gas

Explanation:

There are three (3) states of matter. They are: Solid, Liquid and Gases.

Substance X and Y, belong to the states of matter.

A Solid is a substance that retains its SIZE and SHAPE without need of a container (as opposed to a liquid or gas).

Thus, it will most likely be concluded that: substance X is a Solid; while Y is a Gas

4 0
2 years ago
How many moles of NaOH are in 27.8 mL of 0.168 M NaOH?
Alex17521 [72]
Your answer would be 0.00285 moles.
8 0
3 years ago
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