Express the following numbers in Standard Form:

Which is more massive, the nucleus OR empty space?

pantera1 [17]

Answer:

Empty space

Explanation:

Empty space could be endless sometimes.

8 0

2 years ago

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 is closest packing? What is the difference between hexagonal closest packing and cubic closest packing? What is the unit ce

Paladinen [302]

Answer:

In crystal structure close packing is define as space efficient arrangement of constituent particles to form a crystal lattice.

Explanation:

closest packing

In crystal structure close packing is define as space efficient arrangement of constituent particles to form a crystal lattice.

Difference between hexagonal closest packing and cubic closest packing

In cubic closest packing arrangement, each sphere is surrounded by 12 other spheres on the other hand in case of hexagonal close packing, layers of spheres are packed so that spheres in alternating layers overlie one another.

unit cell for each closest packing arrangement

The hexagonal closest packed arrangement has a coordination number of 12 and it consists of 6 atoms per unit cell. The face centered cubic lattice has a coordination number of 12 and it consists of 4 atoms per unit cell. In case of body centered cubic crystal the coordination number is 8 and it consists of 2 atoms per unit cell.

8 0

3 years ago

How many atoms will there be in 5.00 g of gold. (Use ^ to show the exponent in scientific notation. For example, 4 X 1011 should

gayaneshka [121]

Answer:

1.53x10^22 atoms of Au

Explanation:

To find the atoms of gold we need first, to convert the mass of gold to moles using molar mass of gold (196.97g/mol). Then, these moles must be converted to number of atoms based on definition of moles (1 mole = 6.022x10²³ atoms).

<em>Moles Au:</em>

5.00g Au * (1mol / 196.97g) = 0.0254 moles of Au

<em>Atoms of Au:</em>

0.0254 moles * (6.022x10²³ atoms / 1 mole) =

<h3>1.53x10^22 atoms of Au</h3>

3 0

3 years ago

A sample of carbon dioxide is contained in a 250.0 mL flask at 0.930 atm and 15.4 °C. How many molecules of gas are in

Yanka [14]

Answer:

We are given:

Volume (V) = 0.25 L

Pressure (P) = 0.93 atm

Temperature (T) = 15.4°C OR 288.4 K

<u>Solving for the number of moles of CO₂:</u>

From the ideal gas equation:

PV = nRT

replacing the variables

0.93 * 0.25 = n (0.082)(288.4)

n = 0.00983 moles

<u>Number of molecules:</u>

Number of moles= 0.00983

number of molecules in 1 mole = 6.022 * 10²³

Number of molecules in 0.00983 moles = 0.00983 * 6.022 * 10²³

Number of molecules = 5.91 * 10²¹

8 0

2 years ago