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3 years ago

Question 8 (9 points) Match the lab equipment with its purpose

Chemistry

1 answer:

Deffense [45]3 years ago

7 0

Answer:

$\sf \boxed{4} \mapsto Pipet$

$\sf\boxed{7}\mapsto Test \:tube \: rack$

$\sf\boxed{3}\mapsto Test\: table$

$\sf\boxed{5}\mapsto Scoopula$

$\sf\boxed{1}\mapsto Graduated\: cylinder$

$\sf\boxed{9}\mapsto Bunsen \:burner$

$\sf\boxed{2}\mapsto Beaker$

$\sf \boxed{8}\mapsto Spot\: plate$

$\sf\boxed{6}\mapsto Goggles$

Explanation:

Pipet is used to dispense a very small amount of liquid.

Test tube rack is used to hold multiple test tubes at the same time.

Test Table is used to view chemical reactions or hold or heat small amounts of substance.

Scoopula is used to dispense chemicals from a larger container.

Graduated cylinder is used to measure volume very precisely.

Bunsen burner is used to heat objects.

Beaker is used to transport heat or store substance.

Spot plate is used to observe the color changes of small quantities of a reacting mixture.

Goggles are used to protect the eyes from flying objects or chemical splashes.

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Nicotine, a component of tobacco, is composed of C, H, and N. A 7.875-mg sample of nicotine was combusted, producing 21.363 mg o

Gnom [1K]

Answer: The empirical formula for the given compound is $C_5H_7N$

Explanation:

The chemical equation for the combustion of compound having carbon, hydrogen, and nitrogen follows:

$C_xH_yN_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and nitrogen respectively.

We are given:

Mass of $CO_2=21.363mg=21.363\times 10^3g=21363g$

Mass of $H_2O=6.125g=6.125\times 10^3g=6125g$

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 21363 g of carbon dioxide, $\frac{12}{44}\times 21363=5826.27g$ of carbon will be contained.

For calculating the mass of hydrogen:

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

So, in 6125 g of water, $\frac{2}{18}\times 6125=680.55$ of hydrogen will be contained.

Now we have to calculate the mass of nitrogen.

Mass of nitrogen in the compound = (7875) - (5826.27 + 680.55) = 1368.18 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{5826.27g}{12g/mole}=485.52moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{680.55g}{1g/mole}=680.55moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{1368.18g}{14g/mole}=97.73moles$

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 0.0154 moles.

For Carbon = $\frac{485.52}{97.73}=4.96\approx 5$

For Hydrogen = $\frac{680.55}{97.73}=6.96\approx 7$

For Nitrogen = $\frac{97.73}{97.73}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

Hence, the empirical formula for the given compound nicotine is $C_5H_7N_1=C_5H_7N$

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Answer:

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nydimaria [60]

Hey there!

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WhereasCalcium,Neon have 2 and chlorine has 7.

Hope this explanation helps you :)
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9 moles of reactants chemically change into 11 moles of product.

Explanation:

In the reaction above, 9 moles of reactants chemically change into 11 mole of products. The coefficients in a reaction is the number of moles of the reacting atoms .

For example 8O₂ depicts 8 moles of two oxygen atoms.
The number of moles is a unit for quantifying particles.
You can liken it to a dozen, gross or a score.
Since a mole of a substance contains avogadro number of particles. We can relate the number of moles to other parameters.

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