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3 years ago

Butane methane ethane propane

Chemistry

1 answer:

aleksklad [387]3 years ago

7 0

<h3>Further explanation</h3>

Given

the structure of the alkanes

Required

match the image to the name

Solution

Alkanes are saturated hydrocarbons with all of the carbon bonds being single bonds

General formula:

$\large {\boxed {\bold {C_nH_2n + 2}}}$

If we look at the picture, we define the black atom as Carbon and the branch is Hydrogen

To determine the name of the alkane, all we have to do is count the number of Carbon atoms

Butane : C₄H₁₀ : 4 C atoms

Methane : CH₄ : 1 C atom

Ethane : C₂H₆ : 2 C atoms

Propane : C₃H₈ : 3 C atoms

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Answer:

In the reaction between p-aminophenol and acetic anhydride to form acetaminophen, 4.5 ml of water were added. This was done to recover the product out the container. Water was used as a means of carrying out the product out the container. However, it should be noted that we would want a small amount to be added in order to avoid the product dissolving onto it.

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The half life of plutonium-239 (239pu, pu-239) is 24,100 years. How much plutonium will remain after 1000 years if the initial a

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The mass of plutonium that will remain after 1000 years if the initial amount is 5 g when the half life of plutonium-239 (239pu, pu-239) is 24,100 years is 2.5 g

The equation is Mr=Mi(1/2)^n

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To find n, the number of half-lives, divide the total time 1000 by the time of the half-life(24,100)

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1 year ago

If 16.9 kg of Al2O3(s), 57.4 kg of NaOH(l), and 57.4 kg of HF(g) react completely, how many kilograms of cryolite will be produc

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Answer: 69.72 kg of cryolite will be produced.

Explanation:

The balanced chemical equation is:

$Al_2O_3(s)+6NaOH(l)+12HF(g)\rightarrow 2Na_3AlF_6+9H_2O$

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

moles of $Al_2O_3$ = $\frac{16.9\times 1000g}{102g/mol}=165.7moles$

moles of $NaOH$ = $\frac{57.4\times 1000g}{40g/mol}=1435moles$

moles of $HF$ = $\frac{57.4\times 1000g}{20g/mol}=2870moles$

As 1 mole of $Al_2O_3$ reacts with 6 moles of $NaOH$

166 moles of $Al_2O_3$ reacts with = $\frac{6}{1}\times 166=996$ moles of $NaOH$

As 1 mole of $Al_2O_3$ reacts with 12 moles of $HF$

166 moles of $Al_2O_3$ reacts with = $\frac{12}{1}\times 166=1992$ moles of $HF$

Thus $Al_2O_3$ is the limiting reagent.

As 1 mole of $Al_2O_3$ produces = 2 moles of cryolite

166 moles of $Al_2O_3$ reacts with = $\frac{2}{1}\times 166=332$ moles of cryolite

Mass of cryolite $(Na_3AlF_6)$ = $moles\times {\text {molar mass}}=332mol\times 210g/mol=69720g=69.72kg$

Thus 69.72 kg of cryolite will be produced.

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3 years ago

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