What type of alloy will C and Fe form?

Based on the molecular structures of chloromethane and methane shown, a student makes the claim that a pure sample of chlorometh

goblinko [34]

Answer:

Yes, chloromethane has stronger intermolecular forces than a pure sample of methane has.

Explanation:

In both methane and chloromethane, there are weak dispersion forces. However, in methane, the dispersion forces are the only intermolecular forces present. Also, the lower molar mass of methane means that it has a lower degree of dispersion forces.

For chloromethane, there is in addition to dispersion forces, dipole-dipole interaction arising from the polar C-Cl bond in the molecule. Also the molar mass of chloromethane is greater than that of methane implying a greater magnitude of dispersion forces in operation.

Therefore, chloromethane has stronger intermolecular forces than a pure sample of methane has.

7 0

3 years ago

Text 8.7 Using the virial equation of state for hydrogen at 298 K given in problem 7 (text 8.6), calculate a. The fugacity of hy

gregori [183]

Answer:

a). P = 688 atm

b). P = 1083.04 atm

c).Δ G = 16.188 J/mol

Explanation:

a). Fugacity 'f' can be calculated from the following equations :

$\ln \frac{f}{P}=\int^P_0\left(\frac{Z-1}{P}\right) dP$

where, P = pressure , Z = compressibility

Now, the virial equation is :

$PV=R(1+6.4\times 10^{-4} P)$ ........(1)

Also, PV=ZRT for real gases .......(2)

∴ $ZRT=RT(1+6.4\times 10^{-4} P)$

$Z=1+6.4\times 10^{-4} P$

So from the fugacity equation ,

$\ln \frac{f}{P}=\int^P_0\left(\frac{1+6.4\times 10^{-4}P-1}{P}\right) dP$

$\ln \frac{f}{P}=\int^P_0\left(\frac{6.4\times 10^{-4}P}{P}\right) dP$

$\ln \frac{f}{P} = 6.4 \times 10^{-4} P$

$f=Pe^{6.4 \times 10^{-4} P}$

Putting the value of P = 500 atm in the above equation, we get,

f = 688 atm

b). Given f = 2P

$2P=PE^{6.4 \times 10^{-4}P}$

$2=E^{6.4 \times 10^{-4}P}$

$\ln 2 = 6.4 \times 10^{-4}P$

∴ P = 1083.04 atm

c). dG = Vdp -S dt at constant temperature, dT = 0

Therefore, dG = V dp

$\int^{G_2}_{G_1}dG =\int^{P_2}_{P_1}V dp$

$=\int^{P_2}_{P_1}\frac{RT}{P}\left(1+6.4 \times 10^{-4}P\right) dP$

$=\int^{P_2}_{P_1}RT\frac{dp}{P}+\int^{P_2}_{P_1}RT6.4 \times 10^{-4} dP$

$\Delta G=R[\ln\frac{P_2}{P_1}+6.4 \times 10^{-4}(P_2-P_1)]$

$\Delta G=8.314\times 298[\ln\frac{500}{1}+6.4 \times 10^{-4}(500-1)]$

Δ G = 16.188 J/mol

7 0

3 years ago

What is the mass of carbon in 290 g of CO2?

Elis [28]

Answer:

The mass of CO22 in total is 264 g.

The atomic mass for C is 12 g / mole.

The molar mass for CO22 is (12 + (2 × 16)) = 44 g / mole.

m C = (12 / 44) × 264 = 72 g

So, there are 72 g of C in 264 g of CO2

Explanation:

there's the answer have a good day.

4 0

3 years ago

Write balanced equations for each of the processes, choosing from the following substances as reactants: BaCl2 O2 H2SO4 HNO3 C2H

marishachu [46]

Answer:

BaCl₂ and H₂SO₄

Explanation:

The reaction is the precipitation of BaSO₄ from solution.

This means that our reactants must contain the Ba ion and SO₄²⁻ ion. The reactants that meet this criteria are the; BaCl₂ aand H₂SO₄

The reaction is given as;

BaCl₂ + H₂SO₄ --> BaSO₄ + 2HCl

8 0

4 years ago

How many grams C3H7OH can be made by reacting with 7.3L of CO2 at STP

Komok [63]

Answer:

6.54g of C3H7OH

Explanation:

Step 1:

Determination of the number of mole of CO2 that occupy 7.3L at stp.

This can be obtained as follow:

1 mole of a gas occupy 22.4L at stp.

Therefore, Xmol of CO2 will occupy 7.3L at stp i.e

Xmol of CO2 = 7.3/22.4

Xmol of CO2 = 0.326 mole.

Therefore, 0.326 mole of CO2 was used in the reaction.

Step 2:

The balanced equation for the reaction. This is given below:

6CO2 + 8H2O —> 2C3H7OH + 9O2

Step 3:

Determination of the number of mole of C3H7OH produced from the reaction. This is illustrated below:

From the balanced equation above,

6 moles of CO2 reacted to produce 2 moles of C3H7OH.

Therefore, 0.326 mole of CO2 will react to produce = (0.326 x 2)/6 = 0.109 mole of C3H7OH.

Step 4:

Conversion of 0.109 mole of C3H7OH to grams. This is illustrated below:

Number of mole of C3H7OH = 0.109 mole.

Molar mass of C3H7OH = (12x3)+ (7x1) + 16 + 1 = 60g/mol

Mass of C3H7OH =..?

Mass = mole x molar mass

Mass of C3H7OH = 0.109 x 60

Mass of C3H7OH = 6.54g.

Therefore, 6.54g of C3H7OH is produced from the reaction.

5 0

3 years ago