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3 years ago

Why must pencil be used to draw on paper in paper chromatography

Chemistry

2 answers:

LUCKY_DIMON [66]3 years ago

3 0

Hey there!

The reason why a pencil must be used to draw in a paper chromatography is because its meant to split up the mixture of compounds into its sub parts.

Hope this helped you

<em>Tobey</em>

labwork [276]3 years ago

3 0

Because ink in a pen would be seperated into it's different components on the chromatography paper. Graphite pencils just... don't dissolve like that, they aren't soluble like ink. Graphite would not seperate on the paper, whilst ink would.

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Answer:

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Explanation:

Hello,

In this case, since the reaction between methane and oxygen is:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:

$n_{CH_4}=0.963gCH_4*\frac{1molCH_4}{16gCH_4}=0.0602molCH_4\\ \\n_{CH_4}^{consumed}=7.5gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2} =0.117molCH_4$

Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.

$left\ over=0g$

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2 years ago

hat is the product when magnesium reacts with nitrogen? Mg(s) + N2(g) → Mg2N3(s) Mg3N(s) Mg3N2(s) MgN3(s)

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Answer: The molecular formula for androstenedione is, $C_{19}H_{26}O_2$

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 $CO_2=5.527g$

Mass of $H_2O=1.548g$

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 5.527 g of carbon dioxide, $\frac{12}{44}\times 5.527=1.507g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 1.548 g of water, $\frac{2}{18}\times 1.548=0.172g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.893g)-[(1.507g)+(0.172g)]=0.214g$

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.507g}{12g/mole}=0.126moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.172g}{1g/mole}=0.172moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.214g}{16g/mole}=0.0133moles$

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 $0.0133$ moles.

For Carbon = $\frac{0.126}{0.0133}=9.5$

For Hydrogen = $\frac{0.172}{0.0133}=12.9\approx 13$

For Oxygen = $\frac{0.0133}{0.0133}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 9.5 : 13 : 1

To make in a whole number we are multiplying the ratio by 2, we get:

The ratio of C : H : O = 19 : 26 : 2

Thus, the empirical formula for the given compound is $C_{19}H_{26}O_2$

The empirical formula weight of $C_{19}H_{26}O_2$ = 19(12) + 26(1) + 2(16) = 286 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{286.4}{286}=1$

Molecular formula = $C_{19}H_{26}O_2$

Therefore, the molecular formula for androstenedione is, $C_{19}H_{26}O_2$

3 0

2 years ago