What is the density of a cube that has a mass of 25 g in a volume of 125 cm 3

1 answer:

3 0

<span>The density of an object is defined to be its mass divided by the volume it occupies. For this problem, the mass of the cube was given to be 25 g while its volume is 125 cm</span>³. Thus, we simply divide 25 g by 125 cm³ to get the object’s density. We then calculate that the cube has a density of 0.2 g/ cm³.

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Which of the following is NOT an indication of a chemical reaction?

PolarNik [594]

A. The substance dissolved. Just because a substance dissolved in a solvent doesn’t mean that a chemical action occurred. For example, if you put sugar in water as stir, it will dissolve, but the chemical make up of the sugar and the water don’t change. A chemical reaction can be indicated by a drastic change in color or temperature or if a substance precipitate.

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3 years ago

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Consider this reaction at equilibrium at a total pressure P1: 2SO2(g) + O2(g) â†’ 2SO3(g) Suppose the volume of this system is c

oksian1 [2.3K]

Answer:

The new equilibrium total pressure will be increased to one-half to initial total pressure.

Explanation:

From the information given :

The equation of the reaction can be represented as;

$2SO_{2(g)}+O_{2(g)} \to2SO_{3(g)}$

From above equation:

2 moles of sulphur dioxide reacts with 1 mole of oxygen (i.e 2 moles +1 mole =3 moles ) to give 2 moles of sulphur trioxide

So; suppose the volume of this system is compressed to one-half its initial volume and then equilibrium is reestablished.

So if this process takes place ; the equilibrium will definitely shift to the side with fewer moles , thus the equilibrium will shift to the right. As such; there is increase in pressure.

Let the total pressure at the initial equilibrium be $P_1$

and the total pressure at the final equilibrium be $P_2$

According to Boyle's Law; Boyle's Law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.

Thus;

P ∝ 1/V

P = K/V

PV = K

where K = constant

So;

PV = constant

Hence;

$P_1V_1 = P_2V_2$