All categories

2 years ago

calculate pressure exerted by 1.255 mol of CI2 in a volume of 5.005 L at a temperature 273.5 k using ideal gas equation

Chemistry

1 answer:

balu736 [363]2 years ago

8 0

Answer:

The pressure is 5.62 atm.

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of randomly moving point particles that do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case:

P= ?
V= 5.005 L
n= 1.255 mol
R= 0.082 $\frac{atm*L}{mol*K}$
T= 273.5 K

Replacing:

P* 5.005 L= 1.255 mol* 0.082 $\frac{atm*L}{mol*K}$ *273.5 K

Solving:

$P=\frac{1.255 mol* 0.082 \frac{atm*L}{mol*K} *273.5 K}{5.005 L}$

P= 5.62 atm

The pressure is 5.62 atm.

You might be interested in

A 1000-ml bag of d5w 1/2 ns is to infuse over 8 hours. what is the flow rate of the infusion?

AlladinOne [14]

1542 is the answer. : )

7 0

3 years ago

For the reaction A +B+ C D E, the initial reaction rate was measured for various initial concentrations of reactants. The follow

lora16 [44]

Answer:

Rate constant of the reaction is $3.3\times 10^{-3} M^{-2} s^{-1}$ .

Explanation:

A + B + C → D + E

Let the balanced reaction be ;

aA + bB + cC → dD + eE

Expression of rate law of the reaction will be written as:

$R=k[A]^a[B]^b[C]^c$

Rate(R) of the reaction in trail 1 ,when :

$[A]=0.30 M,[B]=0.30 M,[C]=0.30 M$

$R=9.0\times 10^{-5} M/s$

$9.0\times 10^{-5} M/s=k[0.30 M]^a[0.30 M]^b[0.30 M]^c$ ...[1]

Rate(R) of the reaction in trail 2 ,when :

$[A]=0.30 M,[B]=0.30 M,[C]=0.90 M$

$R=2.7\times 10^{-4} M/s$

$2.7\times 10^{-4} M/s=k[0.30 M]^a[0.30 M]^b[0.90 M]^c$ ...[2]

Rate(R) of the reaction in trail 3 ,when :

$[A]=0.60 M,[B]=0.30 M,[C]=0.30 M$

$R=3.6\times 10^{-4} M/s$

$3.6\times 10^{-4} M/s=k[0.60 M]^a[0.30 M]^b[0.30 M]^c$ ...[3]

Rate(R) of the reaction in trail 4 ,when :

$[A]=0.60 M,[B]=0.60 M,[C]=0.30 M$

$R=3.6\times 10^{-4} M/s$

$3.6\times 10^{-4} M/s=k[0.60 M]^a[0.60 M]^b[0.30 M]^c$ ...[4]

By [1] ÷ [2], we get value of c ;

c = 1

By [3] ÷ [4], we get value of b ;

b = 0

By [2] ÷ [3], we get value of a ;

a = 2

Rate law of reaction is :

$R=k[A]^2[B]^0[C]^1$

Rate constant of the reaction = k

$9.0\times 10^{-5} M/s=k[0.30 M]^2[0.30 M]^0[0.30 M]^1$

$k=\frac{9.0\times 10^{-5} M/s}{[0.30 M]^2[0.30 M]^0[0.30 M]^1}$

$k=3.3\times 10^{-3} M^{-2} s^{-1}$

7 0

3 years ago

Pablo's experiments confirm the relationship between pressure and the number of gas molecules. as the pressure in a system incre

horrorfan [7]

The correct answer is: [C]:
___________________________________________________________
"pressure and the number of gas molecules are directly related."
___________________________________________________________
Note: The conclusion was: " as the pressure in a system increases, the number of gas molecules increases" — over the course of many trials.
This means that the "pressure" and the "number of gas molecules" are directly related.

Furthermore, this conclusion is consistent with the "ideal gas law" equation:

" PV = nRT " ;
____________________________________________________________
in which:

"P = Pressure" ;
"n = number of gas molecules" ;
___________________________________________________________
All other factors held equal, when "n" (the "number of gas molecules")
increases in value (on the "right-hand side" of the equation), the value for "P" (the "pressure" — on the "left-hand side" of the equation), increases.
___________________________________________________________

6 0

3 years ago

Read 2 more answers

A gas is contained in a thick-walled balloon. When the pressure changes from 45.0 bar to 96.0 bar, the volume changes from 1.20

notsponge [240]

P1 * V1 ÷ T1 = P2 * V2 ÷ T2

45 * 1.20 ÷ 314 = 96 * V2 ÷ 420

30,144 * V2 = 22,680

V2 = 22,680 ÷ 30,144

The new volume is approximately 0.75 liter.

I hope I helped

5 0