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2 years ago

All moving objects have kinetic energy. True False

Chemistry

2 answers:

aivan3 [116]2 years ago

4 0

Answer:

true

.......................

nekit [7.7K]2 years ago

4 0

Answer: You're Welcome!

Oh, It's False - Kinetic energy depends upon the speed of the object; potential energy depends upon the position of the object.

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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

3 years ago

What term is used to describe the average kinetic energy of the molecules in a shape?

Bogdan [553]

The term that is used to describe the average kinetic energy of the molecules in a shape is The kinetic theory.

3 0

3 years ago

What is the correct measurement for solubility?

Montano1993 [528]

Other commonly used units include g/L (grams of solute per liter of solution) and m/L (moles of solute per liter of solution). Solubility units always express the maximum amount of solute that will dissolve in either a given amount of solvent, or a given amount of solution, at a specific temperature.

3 0

3 years ago

Calculate the mole fraction of the total ions in an aqueous solution prepared by dissolving 0.400 moles of CaI2 in 850.0 g of wa

Evgesh-ka [11]

Answer:

0.00840

Explanation:

The computation of the mole fraction is as follow:

As we know that

Molar mass = Number of grams ÷ number of moles

number of moles = Number of grams ÷ molar mass

Given that

Number of moles of CaI2 = 0.400

And, Molar mass of water = 18.0 g/mol

Now Number of moles of water is

= 850.0 g ÷ 18.0 g/mol

= 47.22 mol

And, Total number of moles is

= 0.400 + 47.22

= 47.62

So, Molar fraction of CaI2 is

= 0.400 ÷ 47.62

= 0.00840

4 0

3 years ago

. An inflated balloon has a volume of 6.0 L at sea level (1.0 atm) and is allowed to ascend in altitude until the pressure is 0.

Tju [1.3M]

Answer:

The answer to your question is Volume = 11.4 L

Explanation:

Data

Volume 1 = V1 = 6 L

Pressure 1 = P1 = 1 atm

Temperature 1 = T1 = 22°C

Volume 2 = V2 = ?

Pressure 2 = 0.45 atm

Temperature 2 = -21°C

Process

1.- Convert temperature (°C) to °K

T1 = 273 + 22 = 295°K

T2 = 273 + (-21) = 252°K

2.- Use the combined gas law to solve this problem

P1V1 / T1 = P2V2 / T2

-Solve for V2

V2 = P1V1T2 / T1P2

-Substitution

V2 = (6)(1)(252) / (295)(0.45)

- Simplification

V2 = 1512 / 132.75

- Result

V2 = 11.38 L

3 0

2 years ago