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kupik [55]
4 years ago
9

Draw the curved arrows to show the mechanism for the reaction of butanedioic (succinic) anhydride with methanol.

Chemistry
1 answer:
Marta_Voda [28]4 years ago
8 0
Here is the mechanism of said reaction.

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The upper surface of the zone of saturation is called the water table
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Potassium and bromine draw the charges swapping charges formula and name
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I'm not really sure if you're interested in the electron dot diagram of the potassium and bromine atoms, or of potassium bromide,

KBr, so I'll show you both. You can use this example to find the electron dot diagram of hydrogen bromide, HBr.

Explanation:

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An acid-base reaction of (r)-sec-butylamine with a racemic mixture of 2-phenylpropanoic acid forms two products having different
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See explanation

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When an optically active compound reacts with a racemic mixture, two different products appear .

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3 years ago
Which of the following is an exothermic reaction?
Alex_Xolod [135]
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3 years ago
Read 2 more answers
Octane (C8H18) is a component of gasoline. Complete combustion of octane yields H2O and CO2. Incomplete combustion produces H2O
mestny [16]

86.5\; \% of octane had been converted to carbon dioxide CO₂.

<h3>Explanation</h3>

Octane has a molar mass of

12.01 \times 8 + 1.008 \times 18 = 114.22 \; \text{g} \cdot \text{mol}^{-1}

1.000 gallon of this fuel would have a mass of 2.650 kilograms or 2.65 \times 10^{3} \; \text{g}, which corresponds to 2.65 \times 10^{3} / 114.22 = 23.2\; \text{mol} of octane.

Octane undergoes complete combustion to produce carbon dioxide and water by the following equation:

2\; \text{C}_8\text{H}_{18} + 25 \; \text{O}_2 \to 16 \; \text{CO}_2 + 18 \; \text{H}_2\text{O}

An incomplete combustion of octane that gives rise to carbon monoxide and water but no carbon dioxide would consume not as much oxygen:

2\; \text{C}_8\text{H}_{18} + 17 \; \text{O}_2 \to 16 \; \text{CO} + 18 \; \text{H}_2\text{O}

The mass of the product mixture is 11.53 - 2.65 = 8.88 \; \text{kg} heavier than that of the octane supplied. Thus 8.88 \; \text{kg} = 8.88 \times 10^{3} \; \text{g} of oxygen were consumed in the combustion. There are 277.5 \; \text{mol} of oxygen molecules in 8.88 \times 10^{3} \; \text{g} of oxygen.

Let the number of moles of octane that had undergone complete combustion as seen in the first equation be x (0 \le x \le 23.2). The number of moles of octane that had undergone incomplete combustion through the second equation would thus equal 23.2 - x.

25 moles of oxygen gas is consumed for every two moles of octane that had undergone complete combustion and 17 moles if the combustion is incomplete.

n(\text{O}_2, \; \text{Complete Combustion}) + n(\text{O}_2, \; \text{Incomplete Combustion} ) = n(\text{O}_2, \; \text{Consumed})\\

\frac{25}{2} \; x + \frac{17}{2} \; (23.2 - x) = 277.5\\4 \; x = 277.5 - \frac{17}{2} \times 23.2\\x = 20.1

Therefore 20.1 \; \text{mol} out of the 23.2 moles of octane had undergone complete combustion to produce carbon dioxide.

\%n(\text{Complete Combustion}) = 20.1 / 23.2 \times 100 \; \%= 86.5\; \%

7 0
3 years ago
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