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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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How does heat move from one place to another on the troposphere

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3 years ago

What is true about water that is freezing?

Liono4ka [1.6K]

It is going through a physical change

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3 years ago

Read 2 more answers

What assumptions do we make in order to use the Henderson-Hasselbalch equation? a. Both the weak acid and its conjugate base are

zepelin [54]

Answer:

The final and initial concentration of the acid and it's conjugate base are approximately equal, that is we use the weak acid approximation.

Explanation:

The Henderson-Hasselbalch is used to calculate the pH of a buffer solution. It depends on the weak acid approximation.

Since the weak acid ionizes only to a small extent, then we can say that [HA] ≈ [HA]i

Where [HA] = final concentration of the acid and [HA]i = initial concentration of the acid.

It also follows that [A^-] ≈ [A^-]i where [A^-] and[A^-]i refer to final and initial concentrations of the conjugate base hence the answer above.

7 0

3 years ago

Combustion analysis of a 13.42-g sample of the unknown organic compound (which contains only carbon, hydrogen, and oxygen) produ

kirza4 [7]

Answer: The molecular formula for the given organic compound is $C_{18}H_{20}O_2$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=39.61g$

Mass of $H_2O=9.01g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 39.61 g of carbon dioxide, $\frac{12}{44}\times 39.61=10.80g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 9.01 g of water, $\frac{2}{18}\times 9.01=1.00g$ of hydrogen will be contained.

Mass of oxygen in the compound = (13.42) - (10.80 + 1.00) = 1.62 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{10.80g}{12g/mole}=0.9moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1g}{1g/mole}=1moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.62g}{16g/mole}=0.10moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.10 moles.

For Carbon = $\frac{0.9}{0.10}=9$

For Hydrogen = $\frac{1}{0.10}=10$

For Oxygen = $\frac{0.10}{0.10}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 9 : 10 : 1

Hence, the empirical formula for the given compound is $C_9H_{10}O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 268.34 g/mol

Mass of empirical formula = 134 g/mol

Putting values in above equation, we get:

$n=\frac{268.34g/mol}{134g/mol}=2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(9\times 2)}H_{(10\times 2)}O_{(1\times 2)}=C_{18}H_{20}O_2$

Thus, the molecular formula for the given organic compound is $C_{18}H_{20}O_2$ .

3 0

3 years ago

The ph of a 0.55 m aqueous solution of hypobromous acid, hbro, at 25.0 Â°c is 4.48. what is the value of ka for hbro?

gladu [14]

Hbro dissociate as follows
HBro---> H+ + BrO-
Ka= (H+)(BrO-) / HBro
PH = -log (H+)
therefore (H+) = 10^-4.48= 3.31 x10^-5
ka is therefore= ( 3.31 x 10^-5)^2/0.55=1.99 x10^-9

6 0

3 years ago