Assignment

Phoenix [80]

Here's a short answer

Bye Felicia

7 0

3 years ago

Consider the following reaction:C2H4(g) + F2(g) -----------> C2H4F2(g) Delta H = -549 kJEstimate the carbon-fluorine bond ene

frutty [35]

Answer:

Bond energy of carbon-fluorine bond is 485 kJ/mol

Explanation:

Enthalpy change for a reaction, is given as:

$\Delta H_{rxn}=\sum [n_{i}\times (E_{bond})_{i}]-\sum [n_{j}\times (E_{bond})_{j}]$

Where $(E_{bond})_{i}$ and $(E_{bond})_{j}$ represents average bond energy in breaking "i" th bond and forming "j" th bond respectively. $n_{i}$ and $n_{j}$ are number of moles of bond break and form respectively.

In this reaction, one mol of C=C, four moles of C-H and one mol of F-F bonds are broken. One mol of C-C bond, four moles of C-H bonds and two moles of C-F bonds are formed

So, $-549kJ=(1mol\times 614kJ/mo)+(4mol\times E_{C-H})+(1mol\times 154kJ/mol)-(1mol\times 347kJ/mol)-(4mol\times E_{C-H})-(2mol\times E_{C-F})$

or, $-549kJ=(1mol\times 614kJ/mo)+(1mol\times 154kJ/mol)-(1mol\times 347kJ/mol)-(2mol\times E_{C-F})$

or, $E_{C-F}=485kJ/mol$

So bond energy of carbon-fluorine bond is 485 kJ/mol

8 0

3 years ago

What is the de Broglie wavelength, in cm, of a 11.0-g hummingbird flying at 1.20 x 10^2 mph?

KonstantinChe [14]

Answer:1.123 x 10^-31cm

Explanation:

mass of humming bird= 11.0g

speed= 1.20x10^2mph

but I mile = 1.6m

1km=1000

I mile = 1.6x10^3m

1.20x10^2mph= 1.6x10^3m /1mile x at 1.20 x 10^2

=1.932 x10^5m

recall that

1 hr= 60 min

1 min=60 secs, 1hr=3600s

Speed = distance/ time

=1.932 x10^5 / 3600= 5.366 x 10 ^1 m/s

m= a 11.0g= 11.0 x 10^-3kg

h=6.626*10^-34 (kg*m^2)/s

Wavelength = h/mu

= 6.626*10^-34/(11 x 10^-3 x 5.366x 10^1)

6.63x10^-34/ 590.26x 10 ^-3= 1.123 x10^-33m

but 1m = 100cm

1.123 x 10 ^-33 x 100 = 1.123 x 10^-31cm

de broglie wavelength of humming bird = 1.123 x 10 ^-31cm

5 0

3 years ago

A sample of gas occupies 280 mL when the pressure is 560.00 mm Hg . If the temperature remains constant , what is the new pressu

vichka [17]

Answer : The new pressure if the volume changes to 560.0 mL is, 280 mmHg

Explanation :

According to the Boyle's, law, the pressure of the gas is inversely proportional to the volume of gas at constant temperature and moles of gas.

$P\propto \frac{1}{V}$

or,

$P_1V_1=P_2V_2$

where,

$P_1$ = initial pressure = 560.00 mmHg

$P_2$ = final pressure = ?

$V_1$ = initial volume = 280 mL

$V_2$ = final volume = 560.0 mL

Now put all the given values in the above formula, we get:

$560.00mmHg\times 280 mL=P_2\times 560.0 mL$

$P_2=280mmHg$

Therefore, the new pressure if the volume changes to 560.0 mL is, 280 mmHg

3 0

3 years ago

Write the expression for the equilibrium constant Kp for the following reaction. (Enclose pressures in parentheses and do NOT wr

Liono4ka [1.6K]

Answer and Explanation:

For the following balanced reaction:

PCl₅(g) ↔ PCl₃(g) + Cl₂(g)

We can see that all reactants and products are gases, so it is an homogeneous equilibrium. The expression for the equilibrium constant Kp can be written from the partial pressures (P) of reactants and products as follows:

$Kp=\frac{(P PCl_{3})(P Cl_{2})}{(P PCl_{5})}$

Where PPCl₃ is the partial pressure of PCl₃ (reactant), PCl₂ is the partial pressure of Cl₂ (reactant) and PPCl₅ is the partial pressure of PCl₅ (product).

6 0

2 years ago