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3 years ago

Give two real life applications of chromatography

Chemistry

2 answers:

AleksAgata [21]3 years ago

7 0

Answer:

1. Additive detection

2. Crime scene testing

koban [17]3 years ago

4 0

Answer:

Food testing.

Beverage testing.

Drug testing.

Forensic testing.

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can someone explain how this is wrong? I don’t think it is wrong but my chem teacher deducted points. so why?

ruslelena [56]

Answer: measure the mass (48.425g) of KCl

Explanation:

To prepare the solution 0.65M KCl we must measure the mass of KCl that would be dissolved in 1L of the solution. This can be achieved by:

Molar Mass of KCl = 39 + 35.5 = 74.5g/mol

Number of mole (n) = 0.65

Mass conc of KCl = n x molar Mass

Mass conc of KCl = 0.65 x 74.5 = 48.425g

Therefore, to make 0.65M KCl, we must measure 48.425g

7 0

3 years ago

If expending 3500 kcal is equal to a loss of 1.0 lb, how many days will it take Charles to lose 5.0 lb? Express your answer to t

Trava [24]

Answer:

WAIT

good luck on whatever this is for my dude.

3 0

2 years ago

HELPPPP PLEASEE!!!!!!!

OlgaM077 [116]

I think the answer is yes

3 0

3 years ago

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Many double-displacement reactions are enzyme-catalyzed via the "ping pong" mechanism, so called because the reactants appear to

zhenek [66]

Answer:

D. It will decrease by a factor of 4

Explanation:

According to the question , the equation follows :

$A+B\rightarrow C+D$

Rate law : This states the rate of reaction is directly proportional to concentration of reactants with each reactant raised to some power which may or may not be equal to the stoichiometeric coefficient.

$Rate\ \alpha [A]^{a}[B]^{b}$

$r=[A]^{a}[B]^{b}$ .................(1)

STEP": First, find out the power "a" and "b"

a+b = 3 (because it is given that the reaction follow 3rd order-kinetics)

According to question, doubling the concentration of the first reactant causes the rate to increase by a factor of 2 means,

r' = 2r if [A'] = 2[A]

Here [B] is uneffected means [B']=[B]

hence new rate =

$r'=[A']^{a}[B']^{b}$

Put the value of [A'] , [B'] and r' in the above equation:

$2r=[2A]^{a}[B]^{b}$ ...........(2)

Divide equation (1) by (2) we , get

$\frac{2r}{r}=\frac{[2A]^{2}[B]^{b}}{[A]^{a}[B]^{b}}$

$2= 2(\frac{A}{A})^{a}\times (\frac{B}{B})^{b}$

Here A and A cancel each other

B and B cancel each other

We get,

$2= 2^{a}\times 1^{b}$

1^b = 1 ( power of 1 = 1)

$2= 2^{a}$

This is possible only when a = 1

We know that : a + b = 3

1 + b = 3

b =3 -1 = 2

b = 2

Hence the rate law becomes :

$r=[A]^{a}[B]^{b}$

$r=[A]^{1}[B]^{2}$ .............(3)

Look in the question now, it is asked to calculate the concentration of [B],if cut in half

Hence

[B']=1/2[B]

Insert the value of [B'] in equation (3)

$r'=[A]^{1}[B']^{2}$

$r'=[A]^{1}(\frac{1}{2}[B])^{2}$

$r'=\frac{1}{4}[A]^{1}[B]^{2}$ ............(a)

But

$r=[A]^{a}[B]^{b}$ ..............(b)

Compare equation (a) and (b) , we get

new rate r' =

r' = 1/4 r

7 0

3 years ago

Select the correct answer from each drop-down menu.

nataly862011 [7]

The arrow shows that the bond between the chlorine atom and the fluorine atom is nonpolar. The electrons in the bond are pulled more strongly by the fluorine atom, and the chlorine atom is slightly positive.

Explanation:

The bond between Chlorine and fluorine is nonpolar bonding because both of them are sharing an equal number of electrons in the bond. H2, F2, and CL2 are common examples of this.

Chlorine and fluorine are electronegative molecules but Fluorine is above chlorine in the periodic table. Since fluorine is above Chlorine, fluorine has slightly highest electronegative nature compare to fluorine. This is the reason why Fluorine molecules are attracting electrons more than chlorine atoms. This making chlorine atoms slightly positive in Cl and F bonding.

8 0

3 years ago

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