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3 years ago

Which molecule is methylamine?

Chemistry

2 answers:

JulsSmile [24]3 years ago

7 0

Answer:

correct answer is A

Explanation:

Functional group:

The properties of organic compounds are depending on some groups of atom, which is known as functional group. Thus, the compounds, which have same functional group, have similar properties.

For example if the compound contains at least one double bond means it is an example of alkene.

Amines: Amines contain the characteristic –NH2 functional group.

Amine means R-NH2

Methyl amine means CH3NH2

Because meth means one carbon atom only and amine means –NH2 as functional group.

Therefore the correct answer is A

love history [14]3 years ago

3 0

An amine refers to a nitrogen-based compound. Since A is the only option with nitrogen, that would be the answer.

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When a mixture of sulfur and metallic silver is heated, silver sulfide is produced. What mass of silver sulfide is produced from

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The question is incomplete, here is the complete question.

When a mixture of sulfur and metallic silver is heated, silver sulfide is produced. What mass of silver sulfide is produced from a mixture of 3.0g Ag and 3.0g $S_8$ .

Answer : The mass of silver sulfide is produced from a mixture is 3.44 grams.

Explanation : Given,

Mass of Ag = 3.0 g

Mass of $S_8$ = 3.0 g

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First we have to calculate the moles of Ag and $S_8$ .

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$\text{ Moles of }S_8=\frac{\text{ Mass of }S_8}{\text{ Molar mass of }S_8}=\frac{3.0g}{256g/mole}=0.0117moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$16Ag(s)+S_8(s)\rightarrow 8Ag_2S(s)$

From the balanced reaction we conclude that

As, 16 mole of $Ag$ react with 1 mole of $S_8$

So, 0.0278 moles of $Ag$ react with $\frac{0.0278}{16}=0.00174$ moles of $S_8$

From this we conclude that, $S_8$ is an excess reagent because the given moles are greater than the required moles and $Ag$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $Ag_2S$

From the reaction, we conclude that

As, 16 mole of $Ag$ react to give 8 mole of $Ag_2S$

So, 0.0278 moles of $Ag$ react to give $\frac{0.0278}{16}\times 8=0.0139$ moles of $Ag_2S$

Now we have to calculate the mass of $Ag_2S$

$\text{ Mass of }Ag_2S=\text{ Moles of }Ag_2S\times \text{ Molar mass of }Ag_2S$

$\text{ Mass of }Ag_2S=(0.0139moles)\times (247.8g/mole)=3.44g$

Therefore, the mass of silver sulfide is produced from a mixture is 3.44 grams.

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