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hram777 [196]
3 years ago
5

How does the experiment with the marshmallow and the syringe prove that air is matter?

Chemistry
1 answer:
stellarik [79]3 years ago
5 0

solution:

It took mankind tens of thousands of years to figure out that air existed, let alone that it was matter. It was only in recent human history that we figured out anything about air. Proving that air is matter is analogous to today's physics experiments where you cannot see the object of your study, but have to define its properties and its existence from indirect evidence.

We define matter as something which occupies space, is effected by gravity and has weight. Make a vessel that won't collapse if there is no air inside of it. Weigh the vessel when it is full of air. Then pump all of the air out and weigh the vessel again. The difference in weight is the weight of the air.

There is a famous experiment done by Otto Von Guericke in 1654 in Regensburg, Germany. Regensburg was a Roman outpost on the banks of the Danube River. If you ever go there, I highly recommend the Wirstkuke, an 850 year old restaurant near the river. It was there when Guericke was studying air and he may have had a dinner or two there. Anyway, to prove that air exists and has pressure he made a hollow sphere made of two copper halves and sealed it with a gasket. He used an air pump, which he also invented, to pump the air out of the sphere. Air pressure held the two halves of the sphere together. He then took two teams of horses and had them try to pull the sphere apart. They failed. Guericke then opened a valve that let the air back in and that is when the sphere fell apart under its own weight. The sphere was 14" in diameter meaning the air pressure exerted a force of approximately 4.5 tons.

The force would have been the same if one side of the sphere was attached to something fixed, like a really big rock, instead of another team of horses. Guericke might not have understood that or he might have just appreciated the drama of using two teams. Showmanship, you see, is important even in science.

The two most common types of radioactive decay involve the production of alpha particles or beta particles.  

When alpha decay occurs, a daughter element with an atomic number two less than the parent is produced. This is due to the fact that an alpha particle (He nucleus, 2 protons and 2 neutrons) carries away two protons, and the atomic number is the number of protons. In addition, since the alpha particle has four nucleons the mass number of the daughter element goes down by 4.  


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A balloon occupies a volume of 2.00 l at 40.0oc. how much volume will it occupy at 30.0oc?
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<span>Charles' law says "at a constant pressure, the volume of a fixed amount of gas is directly proportional to its absolute temperature".

V </span>α T

Where V is the volume and T is the temperature in Kelvin of the gas. We can use this for two situations as,
V₁/T₁ = V₂/T₂

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By applying the formula,
2.00 L / 313 K = V₂ / 303 K                   
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Use the given data at 500 K to calculate ΔG°for the reaction
Anton [14]

Answer : The  value of \Delta G^o for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0 = standard enthalpy of formation

Now put all the given values in this expression, we get:

\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]

\Delta H^o=-1035.6kJ=-1035600J

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{reactant}

\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S_f^0 = standard entropy of formation

Now put all the given values in this expression, we get:

\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]

\Delta S^o=-153J/K

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 500 K.

\Delta G^o=(-1035600J)-(500K\times -153J/K)

\Delta G^o=-959100J=-959.1kJ

Therefore, the value of \Delta G^o for the reaction is -959.1 kJ

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