Answer:-
2328.454 grams
Explanation:-
Volume V = 18.4 litres
Temperature T = 15 C + 273 = 288 K
Pressure P = 1.5 x 10^ 3 KPa
We know universal Gas constant R = 8.314 L KPa K-1 mol-1
Using the relation PV = nRT
Number of moles of oxygen gas n = PV / RT
Plugging in the values
n = (1.5 x 10^3 KPa ) x ( 18.4 litres ) / ( 8.314 L KPa K-1 mol-1 x 288 K)
n = 11.527 mol
Now the balanced chemical equation for this reaction is
2KNO3 --> 2KNO2 + O2
From the equation we can see that
1 mol of O2 is produced from 2 mol of KNO3.
∴ 11.527 mol of O2 is produced from 2 x 11.527 mol of KNO3.
= 23.054 mol of KNO3
Molar mass of KNO3 = 39 x 1 + 14 x 1 + 16 x 3 = 101 grams / mol
Mass of KNO3 = 23.054 mol x 101 gram / mol
= 2328.454 grams
The volume at STP : 9.856 L
<h3>Further explanation</h3>
Given
Mass of ethane : 13.21 g
Required
The volume at STP
Solution
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.
mol ethane(C2H6) :
= mass : molar mass
= 13.21 g : 30 g/mol
= 0.44
Volume at STP :
= 0.44 x 22.4 L
= 9.856 L
The same sample of gas at different temperatures shows that at low
temperatures, most molecules have speeds close to their average
speed.
<h3>
What does the Maxwell-Boltzmann distribution graph show?</h3>
Put simply, a Maxwell-Boltzmann distribution graph shows how the energy of gas particles varies within a system.
This is solely a measurement of the speeds of particles because kinetic energy is directly related to speed.
The Maxwell-Boltzmann distribution in chemistry is the subject of this article.
We will begin by describing how to read a graph of the Maxwell-Boltzmann distribution. This will involve taking a closer look at things like the typical energy and the most likely energy.
The graph will then be changed under various circumstances, such as when a catalyst is added or the temperature is raised.
The Maxwell-Boltzmann distribution, which we previously mentioned, is a probability function that depicts the distribution of energy among the particles of an ideal gas. (For more information on this topic, see Chemical Kinetics.)
To learn more about Maxwell distribution, refer
to brainly.com/question/24419453
#SPJ4
The temperature at which the process be spontaneous is calculated as follows
delta G = delta H -T delta S
let delta G be =0
therefore delta H- T delta s =0
therefore T= delta H/ delta S
convert 31 Kj to J = 31 x1000= 31000 j/mol
T=31000j/mol /93 j/mol.k =333.33K
