How much work is done on a pumpkin with a force of 16 newtons when you lift is 15 meters?

The two common chlorides of phosphorus, PCl3, and PCl5, both important for the production of the other phosphorus compounds, coe

Irina18 [472]

Answer:

For a: The value of $K_c$ for the given reaction is 271.6

For b: The value of $K_p$ for the reaction is 6.32

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

For $PCl_5$ :

Given mass of $PCl_5$ = 0.105 g

Molar mass of $PCl_5$ = 208.24 g/mol

Putting values in equation 1, we get:

$\text{Moles of }PCl_5=\frac{0.105g}{208.24g/mol}=5.04\times 10^{-4}mol$

For $PCl_3$ :

Given mass of $PCl_3$ = 0.220 g

Molar mass of $PCl_5$ = 137.33 g/mol

Putting values in equation 1, we get:

$\text{Moles of }PCl_3=\frac{0.220g}{137.33g/mol}=1.60\times 10^{-3}mol$

For $Cl_2$ :

Given mass of $Cl_2$ = 2.12 g

Molar mass of $Cl_2$ = 71.0 g/mol

Putting values in equation 1, we get:

$\text{Moles of }Cl_2=\frac{2.12g}{71.0g/mol}=0.029mol$

Volume of the flask = 25.0 L

For the given chemical equation:

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

For a:

The equation used to calculate concentration of a solution is:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}$

The expression of $K_c$ for above reaction follows:

$K_c=\frac{PCl_5}{PCl_3\times Cl_2}$

We are given:

$[PCl_5]=\frac{5.04\times 10^{-4}mol}{25L}$

$[PCl_3]=\frac{1.60\times 10^{-3}mol}{25L}$

$[Cl_2]=\frac{0.029mol}{25L}$

Putting values in above equation, we get:

$K_c=\frac{(\frac{5.04\times 10^{-4}}{25})}{(\frac{1.60\times 10^{-3}}{25})\times (\frac{0.029}{25})}\\\\K_c=271.6$

Hence, the value of $K_c$ for the given reaction is 271.6

For b:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = ?

$K_c$ = equilibrium constant in terms of concentration = 271.6

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = $250^oC=250+273=523K$

$\Delta n_g$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=1-2=-1$

Putting values in above equation, we get:

$K_p=271.6\times (0.0821\times 523)^{-1}\\\\K_p=6.32$

Hence, the value of $K_p$ for the reaction is 6.32

7 0

4 years ago

423 Quiz Work Simple Machines and Power

g100num [7]

I think i think i think i think i think it’s c

7 0

3 years ago

How many mL of 1.76 M sodium hydroxide solution (strong base) must be added to 0.75

erma4kov [3.2K]

Answer: Special care is required to prepare a solution of sodium hydroxide or NaOH in water because considerable heat is liberated by the exothermic reaction. The solution may splatter or boil. Here is how to make a sodium hydroxide solution safely, along with recipes for several common concentrations of NaOH solution.

Explanation:

Hope this helps!

5 0

3 years ago

3. A PS6 plugged into a 120 V outlet uses 3 Amps. What is the Resistance rating of

blondinia [14]

Answer:

R = 40 ohms

Explanation:

Given that,

The voltage of outlet, V = 120 V

Current flowing through the device, I = 3 A

We need to find the Resistance rating of the power cord. Let the resistance is R. We know that,

Ohm's law, V = IR

Put all the values,

$R=\dfrac{V}{I}\\\\R=\dfrac{120}{3}\\\\R=40\ \Omega$

So, the resistance rating of the power cord is equal to 40 ohms.

8 0

3 years ago

Why is a phenol more acidic than a regular alcohol? Group of answer choices

Aleks [24]

Answer:

a.Phenols have the ability to spread out the negative charge that forms.

Explanation:

This happens because of the aromatic ring the phenols have. When loosing the proton, the anion formed have different resonance isomers due to the double bonds in the aromatic ring. This resonance makes the anion more stable and prevents the reaction going backwards.

On the other hand, regular alcohol doesn't have resonace and the reaction of loosing the proton goes leftward and righward: is in equilibrium.

That equilibrium decreases the acidity of the alcohol.

3 0

4 years ago