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m_a_m_a [10]
3 years ago
14

For a successful TLC experiment, you need to pay attention to the details. From the list of instructions and tips below, select

the detail that is NOT important to a TLC experiment. O O Do not disturb the developing chamber. Do not move it or carry around. Closely monitor the temperature of the solvent in the chamber. Use only a plain pencil to draw a starting line and mark the compounds' spots. A pen or marker should never be used. Cover the developing chamber with a watch glass. Ensure the solvent level in the chamber is lower than the starting line where you spot the samples.
Chemistry
1 answer:
Juli2301 [7.4K]3 years ago
5 0

Answer:

Temperature is not that important in TLC.

Explanation:

The concept of TLC is that a solvent (or mixture of solvents) ascends in a capilary system and the differences in interections between solvent and capilary of the substances present allow their separation.

In this process it's important that the air is saturated with solvent vapor, so the lid is important; the solvent must ascend by the capilar, so making it stedy is important, also making sure the starting line is above the solvent level is crucial for this same reason; using a marker would ruin a experiment because most marker's ink (if not all of them) would be carried by the solvent as well, but carbon is not, so a pencil must be used.

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When 6.040 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 18.95 grams of CO2 and 7.759 grams of H
algol [13]

Answer: The empirical formula is CH_2 and molecular formula is C_4H_8

Explanation:

We are given:

Mass of CO_2 = 18.95 g

Mass of H_2O= 7.759 g

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Molar mass of water = 18 g/mol

For calculating the mass of carbon:

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

So, in 18.59 g of carbon dioxide, =\frac{12}{44}\times 18.59=5.07g of carbon will be contained.

For calculating the mass of hydrogen:

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

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Mass of C = 5.07 g

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Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{5.07g}{12g/mole}=0.422moles

Moles of H=\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.862g}{1g/mole}=0.862moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =\frac{0.422}{0.422}=1

For H =\frac{0.862}{0.422}=2

The ratio of C : H = 1: 2

Hence the empirical formula is CH_2.

The empirical weight of CH_2 = 1(12)+2(1)= 14 g.

The molecular weight = 56.1 g/mole

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{56.1}{14}=4

The molecular formula will be=4\times CH_2=C_4H_8

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