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Kaylis [27]
2 years ago
8

What is the name of this branched alkane?

Chemistry
2 answers:
Vlada [557]2 years ago
6 0

Answer:

the answer is methyl pentane

serious [3.7K]2 years ago
4 0

Answer:

Explanation:

3-methyl pentane.

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Which of the following is always a reactant in a combustion reaction? (5 points)
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Answer:

A

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Combustion reaction always has something hydrocarbon plus oxygen gas which equals to carbon dioxide and water. Always being a reactant in a combustion reaction is oxygen, since reactants are what we start off the reaction with.

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A balloon at 32 °C is filled with 21 L of air. What would its volume be at a temperature of 52 °C, assuming pressure remains con
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Answer:C). 24L

Explanation: it works

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When you add an inert gas, the reaction
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Answer: When an inert gas is added to the system in equilibrium at constant pressure, then the total volume will increase.

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Please help do not understand!!!!!!!!!!!!!!!
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2 years ago
In the equation below
yulyashka [42]

Answer:

6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.

Explanation:

We are given the chemical equation:

\displaystyle 2\text{NH$_3$}_\text{(g)} + 3\text{CuO}_\text{(s)} \longrightarrow \text{N$_2$}_\text{(g)}  + 3\text{Cu}_\text{(s)}+3\text{H$_2$O}_\text{(g)}

And we want to determine the amount of products produced when 12.5 moles of NH₃ is reacted with excess CuO.

Compute using stoichiometry. From the equation, we can see the following stoichiometric ratios:

  • The ratio between NH₃ and N₂ is 2:1. (i.e. One mole of N₂ is produced from every two moles of NH₃.)
  • The ratio between NH₃ and Cu is 2:3.
  • The ratio between NH₃ and H₂O is 2:3. (i.e. Three moles of H₂O or Cu is produced frome every two moles of NH₃.)

Dimensional Analysis:

  • The amount of N₂ produced:

\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{1\text{ mol N$_2$}}{2\text{ mol NH$_3$}} = 6.25\text{ mol N$_2$}

  • The amount of Cu produced:

\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{3\text{ mol Cu}}{2\text{ mol NH$_3$}} = 18.8\text{ mol Cu}

  • And the amount of H₂O produced:

\displaystyle 12.5\text{ mol NH$_3$} \cdot \frac{3\text{ mol H$_2$O}}{2\text{ mol NH$_3$}} = 18.8\text{ mol H$_2$O}

In conclusion, 6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.

8 0
2 years ago
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