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3 years ago

In an exothermic reaction, an increase in temperature favors the formation of products.

Chemistry

2 answers:

pentagon [3]3 years ago

7 0

False, in an exothermic reaction, an increase in temperature does not favor the formation of products. Instead, it favors the backward reaction. An exothermic reaction is a reaction where energy is transferred from the system out to the environment.

Ymorist [56]3 years ago

4 0

Answer:

The correct answer is option b.

Explanation:

Exothermic reactions are defined as the reactions in which energy of reactants is more than the energy of the products. In these reactions, energy is released by the system. The total enthalpy of the reaction $(\Delta H)$ comes out to be negative.

In these reaction , products are favored when temperature is kept low. And when we allow these reaction to take place at higher temperature they favor reactant side.

Endothermic reactions are defined as the reactions in which energy of products is more than the energy of the reactants. In these reactions, energy is absorbed by the system. The total enthalpy of the reaction $(\Delta H)$ comes out to be positive.

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When dissolved in water, which of the following would conduct electricity the best?

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Charged particles or ionic particles?

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3 years ago

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Answer:

2 NO (g) → N2 (g) + O2 (g)

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2NOCl (g) ⟶ N2 (g) + O2 (g) + Cl2 (g)

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The ΔH for the reaction is -103.4 kJ

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2 years ago

If a pork roast must absorb kJ to fully cook, and if only 10% of the heat produced by the barbecue is actually absorbed by the r

kolezko [41]

Answer:

1,033.56 grams of carbon dioxide was emitted into the atmosphere.

Explanation:

$C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g),\Delta H_{rxn}=-2044 kJ/mol (assuming)$

Energy absorbed by pork,E = $1.6\times 10^3 kJ$ (assuming)

Total energy produced by barbecue = Q

Percentage of energy absorbed by pork = 10%

$10\%=\frac{E}{Q}\times 100$

$Q=\frac{E}{10}\times 100=\frac{1.6\times 10^3 kJ}{10}\times 100=1.6\times 10^4 kJ$

Since, it is a energy produced in order to indicate the direction of heat produced we will use negative sign.

Q = $-1.6\times 10^4 kJ$

Moles of propane burnt to produce Q energy =n

$n\times \Delta H_{rxn}=Q$

$n=\frac{Q}{\Delta H_{rxn}}=\frac{-1.6\times 10^4 kJ}{-2044 kJ/mol}=7.83 mol$

According to reaction , 1 mol of propane gives 3 moles of carbon dioxide. then 7.83 moles of will give:

$\frac{1}{3}\times 7.83 mol=23.49 mol$ carbon dioxide gas.

Mass of 23.49 moles of carbon dioxide gas:

23.49 mol × 44 g/mol =1,033.56 g

1,033.56 grams of carbon dioxide was emitted into the atmosphere.

8 0

2 years ago

Classify each of the following as energy primarily transferred as HEAT or energy primarily transferred as WORK.

IgorLugansk [536]

number 1 is work number 2 is heat

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3 years ago

Heterocyclic aromatic compounds undergo electrophilic aromatic substitution in a similar fashion to that undergone by benzene wi

Readme [11.4K]

Answer:

See explanation

Explanation:

When we talk about electrophilic substitution, we are talking about a substitution reaction in which the attacking agent is an electrophile. The electrophile attacks an electron rich area of a compound during the reaction.

The five membered furan ring is aromatic just as benzene. This aromatic structure is maintained during electrophilic substitution reaction. The attack of the electrophile generates a resonance stabilized intermediate whose canonical structures have been shown in the image attached.

8 0

3 years ago