Answer:
V = 12.93 L
Explanation:
Given data:
Number of moles = 0.785 mol
Pressure of balloon = 1.5 atm
Temperature = 301 K
Volume of balloon = ?
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Now we will put the values.
V = nRT/P
V = 0.785 mol × 0.0821 atm.L/ mol.K × 301 K / 1.5 atm
V = 19.4 L /1.5
V = 12.93 L
Answer: biology, climate Science, astronomy etc.
Explanation:
The scientific disciplines that are related to chemistry mentioned or alluded to in the article are:
• Biology: Biology is a natural science which studies living organisms, and this include their molecular interactions, evolution, physical structure, and chemical processes.
• Astronomy: Astronomy is the study of space, and the universe. In astronomy, the stars, planets, and galaxies are all studied.
• Climate science: Climate science is also referred to as climatology and it's the scientific study of climate.
Do you have a screen shot or picture of the problem?
Answer:
Entropy increases
Explanation:
Entropy (S) is a measure of the degree of disorder. For a given substance - say water - across phases the following is true ...
S(ice) < S(water) << S(steam)
For a chemical process, entropy changes can be related to increasing or decreasing molar volumes of gas from reactant side of equation to product side of equation. That is ...
if molar volumes of gas increase, then entropy increases, and
if molar volumes of gas decrease, then entropy decreases.
For the reaction 2KClO₃(s) => 2KCl(s) + 3O₂(g)
molar volumes of gas => 0Vm* 0Vm 3Vm
*molar volumes (Vm) apply only to gas phase substances. Solids and liquids do not have molar volume.
Since the reaction produces 3 molar volumes of O₂(g) product vs 0 molar volumes of reactant, then the reaction is showing an increase in molar volumes of gas phase substances and its entropy is therefore increasing.
Answer:
4 monochlorination products can be formed.
Explanation:
Constitutional isomers : These are those compounds with same molecular formula but different atomic arrangement.For example: butane and 2-methly-propane.
On monochlorination of 2 methyl-butane we will have four possibilities of product which will be constitutional isomers of each other:
- 1-chloro-3-methyl butane :
- 2-chloro-3-methyl butane :
- 2-chloro-2-methyl butane
- 1-chloro-2-methyl butane
