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OLga [1]
3 years ago
7

When the products of a reaction are hotter than the reactants: The reaction is exothermic. The reaction is endothermic. The reac

tants lost internal energy. The change in enthalpy is positive.
Chemistry
2 answers:
vodomira [7]3 years ago
6 0

Answer: Option (a) is the correct answer.

Explanation:

A chemical reaction in which energy is released is known as an exothermic reaction.

For example, A + B \rightarrow CD + Heat

In an exothermic reaction, energy of products is less than the energy of reactants. As heat is released that is why, formation of products become hotter.

On the other hand, a chemical reaction in which energy is absorbed by reactant molecules is known as an endothermic reaction.

For example, A + B + Heat \rightarrow CD

In an endothermic reaction, energy of products is more than the energy of reactants. As heat is absorbed, so formation of products become colder.

Thus, we can conclude that when the products of a reaction are hotter than the reactants the reaction is exothermic.

AlekseyPX3 years ago
3 0
<span>Exothermic reaction evolves energy due to which products get hot...</span>
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What is the volume of 0.80 grams of o2 gas at stp? (5 points) group of answer choices 0.59 liters 0.56 liters 0.50 liters 0.47 l
Vladimir [108]

Answer:

0.56L

Explanation:

This question requires the Ideal Gas Law:  PV=nRT where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the Ideal Gas constant, and T is the Temperature of the gas.

Since all of the answer choices are given in units of Liters, it will be convenient to use a value for R that contains "Liters" in its units:R=0.0821\frac{L\cdot atm}{mol\cdot K}

Since the conditions are stated to be STP, we must remember that STP is Standard Temperature Pressure, which means T=273.15K and P=1atm

Lastly, we must calculate the number of moles of O_2(g) there are.  Given 0.80g of O_2(g), we will need to convert with the molar mass of O_2(g).  Noting that there are 2 oxygen atoms, we find the atomic mass of O from the periodic table (16g/mol) and multiply by 2:  32g\text{ }O_2=1mol\text{ }O_2

Thus, \frac{0.80g \text{ }O_2}{1} \frac{1mol\text{ }O_2}{32g\text{ }O_2}=0.25mol\text{ }O_2=n

Isolating V in the Ideal Gas Law:

PV=nRT

V=\frac{nRT}{P}

...substituting the known values, and simplifying...

V=\frac{(0.025 mol \text{ }O_2)(0.0821\frac{L\cdot atm}{mol \cdot K} )(273.15K)}{(1atm)}

V=0.56L \text{ } O_2

So, 0.80g of O_2(g) would occupy 0.56L at STP.

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Answer:

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