Ask question

Ask question

All categories

2 years ago

6

Density of 30.00 mL of H2O

1 answer:

ohaa [14]2 years ago

6 0

its not possible sorry

You might be interested in

What do you think a chemical change is?

Rudiy27

Answer:

a chemical change is when the substance turns into a new substance and can not be changed back into its original form

Explanation:

6 0

3 years ago

Answer the questions below:

Helga [31]

Why do people always send those links like it’s helpful, i’ll help you!

8 0

2 years ago

Read 2 more answers

An isotope undergoes radioactive decay by emitting radiation that has no mass. What other characteristic does the radiation

geniusboy [140]

Answer:

no charge

Explanation:

answer on edge

8 0

3 years ago

An equilibrium mixture of PCl5(g), PCl3(g), and Cl2(g) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 Torr, respective

djverab [1.8K]

Answer:

The new partial pressures after equilibrium is reestablished for $PCl_3$ :

$P_1'=6.798 Torr$

The new partial pressures after equilibrium is reestablished $Cl_2$ :

$P_2'=26.398 Torr$

The new partial pressures after equilibrium is reestablished for $PCl_5$ :

$P_3'=223.402 Torr$

Explanation:

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

At equilibrium before adding chlorine gas:

Partial pressure of the $PCl_3=P_1=13.2 Torr$

Partial pressure of the $Cl_2=P_2=13.2 Torr$

Partial pressure of the $PCl_5=P_3=217.0 Torr$

The expression of an equilibrium constant is given by :

$K_p=\frac{P_1}{P_1\times P_2}$

$=\frac{217.0 Torr}{13.2 Torr\times 13.2 Torr}=1.245$

At equilibrium after adding chlorine gas:

Partial pressure of the $PCl_3=P_1'=13.2 Torr$

Partial pressure of the $Cl_2=P_2'=?$

Partial pressure of the $PCl_5=P_3'=217.0 Torr$

Total pressure of the system = P = 263.0 Torr

$P=P_1'+P_2'+P_3'$

$263.0Torr=13.2 Torr+P_2'+217.0 Torr$

$P_2'=32.8 Torr$

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

At initail

(13.2) Torr (32.8) Torr (13.2) Torr

At equilbriumm

(13.2-x) Torr (32.8-x) Torr (217.0+x) Torr

$K_p=\frac{P_3'}{P_1'\times P_2'}$

$1.245=\frac{(217.0+x)}{(13.2-x)(32.8-x)}$

Solving for x;

x = 6.402 Torr

The new partial pressures after equilibrium is reestablished for $PCl_3$ :

$P_1'=(13.2-x) Torr=(13.2-6.402) Torr=6.798 Torr$

The new partial pressures after equilibrium is reestablished $Cl_2$ :

$P_2'=(32.8-x) Torr=(32.8-6.402) Torr=26.398 Torr$

The new partial pressures after equilibrium is reestablished for $PCl_5$ :

$P_3'=(217.0+x) Torr=(217+6.402) Torr=223.402 Torr$

5 0

2 years ago

I turn the Ritz into a poor house

Ket [755]

Low life?- i’m pretty sure

6 0

2 years ago