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1 year ago

How many molecules are there in 79g of Fe2O3? how many atoms is this?

1 answer:

4 0

There are approximately 160 grams in 1 mol of Fe2O3 molecules. Therefore, there would be 79/160= 0.49375 mols of Fe2O3 molecules in 79 grams. There are 5 atoms in total for each molecule of Fe2O3, therefore 79/160 * 5 = 79/32 = 2.46875 mols of atoms.

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A 3.5 kg iron shovel is left outside through the winter. The shovel, now orange with rust, is rediscovered in the spring. Its ma

Anton [14]

Answer:

Explanation:

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1 year ago

At a given temperature, you have a mixture of benzene (vapor pressure of pure benzene = 745 torr) and toluene (vapor pressure of

swat32

Ideal solutions obey Raoult's law, which states that:

P_i = x_i*(P_pure)_i

where
P_i is the partial pressure of component i above a solution
x_i is the mole fraction of component i in the solution
(P_pure)_i is the vapor pressure of pure component i

In this case,

P_benzene = 0.59 * 745 torr = 439.6 torr
P_toluene = (1-0.59) * 290 torr = 118.9 torr

The total vapor pressure above the solution is the sum of the vapor pressures of the individual components:

P_total = (439.6 + 118.9) torr = 558.5 torr

Assuming the gas phase also behaves ideally, the partial pressure of each gas in the vapor phase is proportional to its molar concentration, so the mole fraction of toluene in the vapor phase is:

118.9 torr/558.5 torr = 0.213

8 0

1 year ago

Iron (Fe) undergoes an allotropic transformation at 912°C: upon heating from a BCC (α phase) to an FCC (γ phase). Accompanying t

-Dominant- [34]

Answer:

The description including its given problem is outlined in the following section on the clarification.

Explanation:

The given values are:

RBCC = 0.12584 nm

RFCC = 0.12894 nm

The unit cell edge length (ABCC) as well as the atomic radius (RBcc) respectively connected as measures for BCC (α-phase) structure:

√3 ABCC = 4RBCC

⇒ ABCC = $\frac{4RBCC}{\sqrt{3} }$

⇒ = $\frac{4\times 0.12584}{\sqrt{3}}$

⇒ = $0.29062 \ nm$

Likewise AFCC as well as RFCC are interconnected by

√2AFCC = 4RFCC

⇒ AFCC = $\frac{4RFCC}{\sqrt{2}}$

⇒ = $\frac{4\times 0.12894}{\sqrt{2} }$

⇒ = $0.36470 \ nm$

Now,

The Change in Percent Volume,

= $\frac{V \ final-V \ initial}{V \ initial}\times 100 \ percent$

= $\frac{(VFCC)unit \ cell-(VBCC)unit \ cell}{(VBCC)unit \ cell}\times 100 \ percent$

= $\frac{(aFCC)^3-(aBCC)^3}{(aBCC)^3}\times 100 \ percent$

= $\frac{(0.36470)^3-(0.29062)^3}{(0.29062)^3}\times 100 \ percent$

= $97.62 \ percent (approximately)$

Note: percent = %

7 0

1 year ago

Moving from boron to carbon, the intensity of the bulb __

AleksandrR [38]

Explanation:

The thickness of the frosting Moving from boron to carbon, the intensity of the bulb blank because Z increases from blank to blank. The thickness of the frosting blank because the core electron configuration is the same for both atoms. because the core electron configuration is the same for both atoms.

7 0

1 year ago

Most Bic lighters hold 5.0ml of liquified butane (density = 0.60 g/ml). Calculate the minimum size container you would need to "

Hatshy [7]

Answer:

Volume of container = 0.0012 m³ or 1.2 L or 1200 ml

Explanation:

Volume of butane = 5.0 ml

density = 0.60 g/ml

Room temperature (T) = 293.15 K

Normal pressure (P) = 1 atm = 101,325 pa

Ideal gas constant (R) = 8.3145 J/mole.K)

volume of container V = ?

Solution

To find out the volume of container we use ideal gas equation

PV = nRT

P = pressure

V = volume

n = number of moles

R = gas constant

T = temperature

First we find out number of moles

As Mass = density × volume

mass of butane = 0.60 g/ml ×5.0 ml

mass of butane = 3 g

now find out number of moles (n)

n = mass / molar mass

n = 3 g / 58.12 g/mol

n = 0.05 mol

Now put all values in ideal gas equation

PV = nRt

V = nRT/P

V = (0.05 mol × 8.3145 J/mol.K × 293.15 K) ÷ 101,325 pa

V = 121.87 ÷ 101,325 pa

V = 0.0012 m³ OR 1.2 L OR 1200 ml

8 0

1 year ago