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guajiro [1.7K]
3 years ago
15

Did Chelate compounds increase the rate of reaction?

Chemistry
1 answer:
Viefleur [7K]3 years ago
7 0

Answer:

Chelate, any of a class of coordination or complex compounds consisting of a central metal atom attached to a large molecule, called a ligand, in a cyclic or ring structure. An example of a chelate ring occurs in the ethylenediamine-cadmium complex:

The ethylenediamine ligand has two points of attachment to the cadmium ion, thus forming a ring; it is known as a didentate ligand. (Three ethylenediamine ligands can attach to the Cd2+ ion, each one forming a ring as depicted above.) Ligands that can attach to the same metal ion at two or more points are known as polydentate ligands. All polydentate ligands are chelating agents.

Chelates are more stable than nonchelated compounds of comparable composition, and the more extensive the chelation—that is, the larger the number of ring closures to a metal atom—the more stable the compound. This phenomenon is called the chelate effect; it is generally attributed to an increase in the thermodynamic quantity called entropy that accompanies chelation. The stability of a chelate is also related to the number of atoms in the chelate ring. In general, chelates containing five- or six-membered rings are more stable than chelates with four-, seven-, or eight-membered rings.

Explanation:

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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
An unknown piece of metal weighing 95.0 g is heated to 98.0°C. It is dropped into 250.0 g of water at 23.0°C. When equilibrium i
lutik1710 [3]

Answer:

C_{metal}=126.6\frac{J}{g\°C}

Explanation:

Hello!

In this case, when two substances at different temperature are put in contact and an equilibrium temperature is attained, we can evidence that the heat lost by the hot substance (metal) is gained by the cold substance (water) and we can write:

Q_{metal}=-Q_{water}

Which can be also written as:

m_{metal}C_{metal}(T_{EQ}-T_{metal})=-m_{water}C_{water}(T_{EQ}-T_{water})

Thus, since we need the specific heat of the metal, we solve for it as shown below:

C_{metal}=\frac{m_{water}C_{water}(T_{EQ}-T_{water})}{-m_{metal}(T_{EQ}-T_{metal})} \\\\C_{metal}=\frac{250.0g*4.184\frac{J}{g\°C}(29.0\°C-98.0\°C)}{95.0g(29.0\°C-23.0\°C)} \\\\C_{metal}=126.6\frac{J}{g\°C}

Best regards.

7 0
3 years ago
C6H14 (g) + O2 (g) -&gt; H20 (g) + CO2 (g)<br> chemical reaction?
zysi [14]

Answer:

Single displacement (Substitution)

Explanation:

3 0
3 years ago
Which of the following choices best explains why glaciers move?
irinina [24]
D. Should be correct- the other ones don’t make sense
4 0
3 years ago
Read 2 more answers
A scientist is wondering why a certain region in the ocean doesn't have maximum phytoplankton growth, despite having plenty of n
Advocard [28]

Answer:

C The water had adequate nitrogen and phosphorus, so it is likely iron limited.

Explanation:

Phytoplankton are single- cell organisms that live in oceans.

They require nitrogen, phosphorus and trace amount of iron to survive.

From the scientist's results after testing the water for nitrogen and phosphorus,there are reasonable amount of these elements.

Therefore insufficient iron in the water is the reason why he could find plenty phytoplankton in the ocean.

8 0
3 years ago
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