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Ivahew [28]
3 years ago
15

PLEASE HELP ME I HAVE NO IDEA “Heat (thermal) energy is being applied to the substance whenever the Bunsen Burner is on. Before

the solid begins to melt, how is this energy being used?”
Chemistry
1 answer:
svetlana [45]3 years ago
5 0

TLDR: The energy was being used simply to heat the substance up.

Whenever something melts, it performs what is called a "phase transition", where the state of matter moves from one thing to something else. You can see this in your iced drink at lunch; as the ice in the cup of liquid heats up, it reaches a point where it will eventually "change phase", or melt. The same can be achieved if you heat up that water enough, like if you're cooking; when you boil eggs, the water has so much thermal energy it can "change phase" and become a gas!

However, water doesn't randomly become a boiling gas, it has to heat up for a while before it reaches that temperature. For a real-life example, the next time you cook something, hold you hand above the water before it starts boiling. You'll see that that water has quite a high temperature despite not boiling.

There's a lot of more complex chemistry to describe this phenomena, such as the relationship between the temperature, pressure, and what is called the "vapor pressure" of a liquid when describing phase changes, but for now just focus on the heating effect. When ice melts, it doesn't seem like its heating up, but it is. The ice absorbs energy from its surroundings (the warmer water), thus heating up the ice and cooling down the water. Similarly, the bunsen burner serves to heat up things in the lab, so before the solid melts in this case it was simply heating up the solid to the point that it <u>could</u> melt.

Hope this helps!

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Ethene is converted to ethane by the reaction flows into a catalytic reactor at 25.0 atm and 250.°C with a flow rate of 1050. L/
Sphinxa [80]

Answer : The percent yield of the reaction is, 76.34 %

Explanation : Given,

Pressure of C_2H_4 and H_2 = 25.0 atm

Temperature of C_2H_4 and H_2 = 250^oC=273+250=523K

Volume of C_2H_4 = 1050 L per min

Volume of H_2 = 1550 L per min

R = gas constant = 0.0821 L.atm/mole.K

Molar mass of C_2H_6 = 30 g/mole

First we have to calculate the moles of C_2H_4 and H_2 by using ideal gas equation.

For C_2H_4 :

PV=nRT\\\\n=\frac{PV}{RT}

n=\frac{PV}{RT}=\frac{(25atm)\times (1050L)}{(0.0821L.atm/mole.K)\times (523K)}

n=611.34moles

For H_2 :

PV=nRT\\\\n=\frac{PV}{RT}

n=\frac{PV}{RT}=\frac{(25atm)\times (1550L)}{(0.0821L.atm/mole.K)\times (523K)}

n=902.46moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

C_2H_4+H_2\rightarrow C_2H_6

From the balanced reaction we conclude that

As, 1 mole of C_2H_4 react with 1 mole of H_2

So, 611.34 mole of C_2H_4 react with 611.34 mole of H_2

From this we conclude that, H_2 is an excess reagent because the given moles are greater than the required moles and C_2H_4 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of C_2H_6.

As, 1 mole of C_2H_4 react to give 1 mole of C_2H_6

As, 611.34 mole of C_2H_4 react to give 611.34 mole of C_2H_6

Now we have to calculate the mass of C_2H_6.

\text{Mass of }C_2H_6=\text{Moles of }C_2H_6\times \text{Molar mass of }C_2H_6

\text{Mass of }C_2H_6=(611.34mole)\times (30g/mole)=18340.2g

The theoretical yield of C_2H_6 = 18340.2 g

The actual yield of C_2H_6 = 14.0 kg = 14000 g      (1 kg = 1000 g)

Now we have to calculate the percent yield of C_2H_6

\%\text{ yield of }C_2H_6=\frac{\text{Actual yield of }C_2H_6}{\text{Theoretical yield of }C_2H_6}\times 100=\frac{14000g}{18340.2g}\times 100=76.34\%

Therefore, the percent yield of the reaction is, 76.34 %

5 0
4 years ago
To indicate the pH of a substance, one can use a dye that is both an acid and that appears as different colors in its protonated
PIT_PIT [208]

Answer:

<em>The pH of the solution is 7.8</em>

Explanation:

The concentration of the solution is 0.001M and  the dye could be in its protonated and deprotonated forms. If the concentration of the protonated form [HA] is 0.0002 M the concentration of the deprotonated form will be the subtraction between the concentration of the bye and the concentration of the protonated form:

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Also, the Henderson-Hasselbalch equation is

pH = pKa + Log \frac{[A^{-}]}{[HA]}

this equation shows the dependency between the pH of the solution, the pKa and the concentration of the protonated and deprotonated forms. Thus, replacing in the equation

pH = 7.2 + Log \frac{0.0008M}{0.0002M} = 7.8

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Explanation:

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