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Ivanshal [37]
4 years ago
15

Bob the chemist ran a TLC plate of his product, but the spot would not move off the baseline. The eluent he used consisted of a

30% polar / 70% non-polar solvent system. Which solvent system should he try next in order to get the compounds in his product mixture to move up the plate
Chemistry
1 answer:
olya-2409 [2.1K]4 years ago
7 0

Answer:

40 % polar / 60 % nonpolar system

Explanation:

According to the like-dissolves-like rule, remember that polar compounds dissolve in polar solvents and nonpolar compounds dissolve in nonpolar solvents.

Notice that the eluent used is only slightly polar and mainly nonpolar. Since the product didn't move at all, this means it didn't interact with the major nonpolar solvent and is expected to be polar.

That said, a more polar system should be used in order to make the product move off the baseline, such as 40 % polar and 60 % nonpolar solvent system.

In case the product still doesn't move, polarity of the eluent should be increased further.

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The production of NOx gases is an unwanted side reaction of the main engine combustion process that turns octane, C8H18, into CO
borishaifa [10]

Answer:

710,33 g NO2

Explanation:

2 C8H18 + 25 O2 → 16 CO2 + 18 H2O  

(800 g octane) / (114.2293 g C8H18/mol x (25/2)) = 87.54 mol O2 used to combust the octane

87.54 mol O2 x \frac{15}{85} = 15.44 mol O2 used to form NO2

O2 + 2NO → 2NO2

(15.44 mol O2) x (2/2) x (46.0056 g NO2/mol) = 710,33 g NO2

4 0
3 years ago
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I need help with this question please!!
mariarad [96]
The answer is B

Hope this helped. Good luck!
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3 years ago
How many moles are in 3.0g of A1203?
vfiekz [6]

Answer:

0.065 moles

Explanation:

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3 years ago
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11111nata11111 [884]
The correct answer is B. Homeostasis
5 0
3 years ago
Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and O. A 0.1005 g sample of menthol is comb
Bogdan [553]

<u>Answer:</u> The molecular formula for the menthol is C_{10}H_{20}O

<u>Explanation:</u>

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=0.2829g

Mass of H_2O=0.1159g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

  • <u>For calculating the mass of carbon:</u>

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

So, in 0.2829 g of carbon dioxide, \frac{12}{44}\times 0.2829=0.077g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

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

So, in 0.1159 g of water, \frac{2}{18}\times 0.1159=0.0129g of hydrogen will be contained.

  • Mass of oxygen in the compound = (0.1005) - (0.077 + 0.0129) = 0.0106 g

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

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon = \frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.077g}{12g/mole}=0.0064moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.0129g}{1g/mole}=0.0129moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.0106g}{16g/mole}=0.00066moles

  • <u>Step 2:</u> 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 0.00066 moles.

For Carbon = \frac{0.0064}{0.00066}=9.69\approx 10

For Hydrogen  = \frac{0.0129}{0.00064}=19.54\approx 20

For Oxygen  = \frac{0.00066}{0.00066}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 10 : 20 : 1

Hence, the empirical formula for the given compound is C_{10}H_{20}O_1=C_{10}H_{20}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 = 156.27 g/mol

Mass of empirical formula = 156 g/mol

Putting values in above equation, we get:

n=\frac{156.27g/mol}{156g/mol}=1

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

C_{(1\times 10)}H_{(1\times 20)}O_{(1\times 1)}=C_{10}H_{20}O

Thus, the molecular formula for the menthol is C_{10}H_{20}O

4 0
3 years ago
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