What is the percent by mass of phosphorus?

What volume (in liters, at 703 k and 2.04 atm) of chlorine gas is required to react with 3.39 g of p?

Natali5045456 [20]

The volume of chlorine required is 7.71 L.

The reaction between phosphorus and chlorine is:

2P + 5Cl₂→ 5PCl₅

Therefore, 2 moles of P requires 5 moles of chlorine to react with it.

Given mass of P =3.39 g

Molar mass of P=30.97 g/mol

No. of moles of P = given mass/ molar mass = 3.39 / 30.97 = 0.109 moles

2 moles of P requires 5 moles of chlorine

0.109 moles of P will require 0.109 x 5/2 = 0.2725 moles of chlorine

According to ideal gas equation

PV=nRT

2.04 x V = 0.2725 x 0.0821 x 703

V = 0.2725 x 0.0821 x 703 / 2.04

V = 7.71L

Learn more about ideal gas equation here:

brainly.com/question/3637553

#SPJ4

5 0

2 years ago

Does the model below represent a mixture or a substance ?

Vikentia [17]

Answer:

substance

Explanation:

A mixture is when two or more <u>different</u> atoms/molecules are together, but not joined.

A substance is when the <u>same </u>atom/molecule is in a group together.

In this example, it is a substance because it is comprised of the same molecule not joined all together. If you wanted a mixture, other colored atoms/molecule (e.g. add green atoms) would change it to this property.

4 0

3 years ago

Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and

skelet666 [1.2K]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

Explanation :

The given balanced chemical reaction is,

$C_6H_6(l)\rightarrow C_6H_6(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0_{(C_6H_6(g))}$ = standard enthalpy of formation of gaseous benzene = 82.93 kJ/mol

$\Delta H_f^0_{(C_6H_6(l))}$ = standard enthalpy of formation of liquid benzene = 49.04 kJ/mol

Now put all the given values in this expression, we get:

$\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]$

$\Delta H^o=33.89kJ/mol=33890J/mol$

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S^0_{(C_6H_6(g))}$ = standard entropy of formation of gaseous benzene = 269.2 J/K.mol

$\Delta S^0_{(C_6H_6(l))}$ = standard entropy of formation of liquid benzene = 173.26 J/K.mol

Now put all the given values in this expression, we get:

$\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]$

$\Delta S^o=95.94J/K.mol$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is $25^oC\text{ or }298K$ .

$\Delta G^o=(33890J)-(298K\times 95.94J/K)$

$\Delta G^o=5299.88J/mol=5.299kJ/mol$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

6 0

3 years ago

Read 2 more answers

Que impacto ambiental hay de los servicios de integrales de diseño y contracción

Nuetrik [128]

Answer:

IMPACTO AMBIENTAL EN LOS DISEÑOS Y CONSTRUCCIÓN

Explanation:

Los impactos directos por el uso de las vías terrestres pueden incluir: mayor demanda de combustibles para los motores; accidentes con los medios no motorizados de transporte o el reemplazo de los mismos; mayor contaminación del aire, ruido, desechos a los lados del camino; daños físicos o muerte a animales y personas ...

8 0

3 years ago

Define the term chemotherapy.

Ksivusya [100]

I'm not able to understand that what is written in the pic.. As I can't understand this language..

Treatment that uses drugs to stop the growth of cancer cells, either by killing the cells or by stopping them from dividing. Chemotherapy may be given by mouth, injection, or infusion, or on the skin, depending on the type and stage of the cancer being treated.

8 0

3 years ago

Read 2 more answers