Answer:
Density = mass/volume
= 44/22.4
= 1.96 gram/liter
The density of the Carbon Dioxide at S.T.P. (Standard Temperature and Volume) is 1.96 gram/liter.
Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.
Explanation :
(a) At constant volume condition the entropy change of the gas is:
We know that,
The relation between the 
for an ideal gas are :
As we are given :
Now we have to calculate the entropy change of the gas.
(b) As we know that, the work done for isochoric (constant volume) is equal to zero. 
(C) Heat during the process will be,
Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.
Answer:
M = 20.5 g/mol
Explanation:
Given data:
Volume of gas = 1.20 L
Mass of gas = 1.10 g
Temperature and pressure = standard
Solution:
First of all we will calculate the density.
Formula:
d = mass/ volume
d = 1.10 g/ 1.20 L
d = 0.92 g/L
Now we will calculate the molar mass.
d = PM/RT
0.92 g/L = 1 atm × M / 0.0821 atm.L/mol.K ×273.15 K
M = 0.92 g/L × 0.0821 atm.L/mol.K ×273.15 K / 1 atm
M = 20.5 g/mol
Answer:
Chemical reactions make and break the chemical bonds between molecules, resulting in new materials as the products of the chemical reaction.
Explanation:
Breaking chemical bonds absorbs energy, while making new bonds releases energy, with the overall chemical reaction being endothermic or exothermic.
