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3 years ago

What device is often used to locate primer and gunpowder particles by their characteristic size and shape?

1 answer:

6 0

A chemical test used to develop patterns of gunpowder residues around bullet holes ... -the inner surface of the barrel of a gun leaves its markings on a bullet ... pin will be impressed into the relatively soft metal or the primer on the cartridge case ... such as the size and the shape of the tool (rarely individual characteristics)...

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How many hydrogen atoms are in a cycloalkane with 6 carbon atoms?

Anna007 [38]

A cycloalkane is a regular alkane with a ring or loop. An example is cyclohexane, which is a ring of 6 carbon atoms, each bonded to 2 hydrogen atoms

3 0

4 years ago

What is the name of the molecular geometry (shape) for this Lewis Structure?

Nikitich [7]

Rara tee da goo boo boo teee gaaa baa baa

7 0

4 years ago

Many moderately large organic molecules containing basic nitrogen atoms are not very soluble in water as neutral molecules, but

alukav5142 [94]

Answer:

a) For nicotine, the protonated form is the present in stomach.

b) For caffeine, the neutral base form is the present in stomach.

c) For Strychinene, the protonated form is the present in stomach.

d) For quinine, the protonated form is the present in stomach.

Explanation:

In a basic dissociation for molecules with basic nitrogen, the equilibrium is:

A + H₂O ⇄ AH⁺ + OH⁻

Where A is neutral base and AH⁺ is protonated form

The basic dissociation constant, kb, is:

$K_{b} = \frac{[AH^+][OH^-]}{[A]}$

As pH in stomach is 2,5:

[OH] = $10^{-[14-pH]}$

[OH] = 3,16x10⁻¹² M

Thus:

$\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$ (1)

Using (1) it is possible to know if you have the neutral base or the protonated form, thus:

(a) nicotine Kb = 7x10^-7

$\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$\frac{7x10^{-7}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$221359 = \frac{[AH^+]}{[A]}$

[AH⁺}>>>>[A]

For nicotine, the protonated form is the present in stomach

(b) caffeine,Kb= 4x10^-14

$\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$\frac{4x10^{-14}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$0,0127 = \frac{[AH^+]}{[A]}$

[AH⁺}<<<<[A]

For caffeine, the neutral base form is the present in stomach

(c) strychnine Kb= 1x10^-6

$\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$\frac{1x10^{-6}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$314456 = \frac{[AH^+]}{[A]}$

[AH⁺}>>>>[A]

For Strychinene, the protonated form is the present in stomach

(d) quinine, Kb= 1.1x10^-6

$\frac{K_{b}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$\frac{1,1x10^{-6}}{3,16x10^{-12}} =\frac{[AH^+]}{[A]}$

$347851= \frac{[AH^+]}{[A]}$

[AH⁺}>>>>[A]

For quinine, the protonated form is the present in stomach.

I hope it helps!

7 0

3 years ago

What happens when the concentration of water inside a cell is lower than the concentration of water outside the cell?

adell [148]

Answer:

Water outside the cell will flow inwards by osmosis to attain equilibrium

Explanation:

In the hypotonic environment, the concentration of water is greater outside the cell and the concentration of solute is higher inside. A solution outside of a cell has a lower concentration of solutes relative to the cytosol.

If concentrations of dissolved solutes are greater inside the cell, the concentration of water inside the cell is correspondingly lower. As a result, water outside the cell will flow inwards by osmosis to attain equilibrium.

Osmosis is a process by which molecules of a solvent tend to pass from a less concentrated solution into a more concentrated one through a semipermeable membrane.

4 0

3 years ago

Which statement is incorrect regarding the reaction of benzene with an electrophile? View Available Hint(s) A) The carbocation i

Leno4ka [110]

Answer:

The carbocation intermediate reacts with a nucleophile to form the addition product.

Explanation:

The reaction of benzene with an electrophile is an electrophillic substitution reaction. Here the electrophile replaces hydrogen. There is no formation of carbocation as intermediate in the reaction. Infact there is transition state where the electorphile attacks on benzene ring and at the same time the hydrogen gets removed from the benzene. So a transition carbocation is formed.

The general mechanism is shown in the figure.

i) Attack of the electrophile on the benzene (which is the nucleophile)

ii) The carbocation intermediate loses a proton from the carbon bonded to the electrophile.

iii) the carbocation formation is the rate determining step.

iv) There is no formation of addition product.

Thus the wrong statement is

The carbocation intermediate reacts with a nucleophile to form the addition product.

3 0

4 years ago