The volume of the gas that occupy at STP is 165. 28 cm^3
calculation
by use of combined gas law that is P1V1/T1=P2V2/T2, where
P1=84.6 kpa
T1=23.5 +273=296.5 K
V1=215 cm^3
At STP T= 273 K and P= 101.325 Kpa
therefore p2 = 101.325 Kpa and T2 = 272 K V2=?
by making V2 the subject of the formula V2 =T2P1V1/P2T1
V2 = 273 K x 84.6 Kpa x 215 cm^3/ 101,.325 Kpa x296.5 K =165.28 cm^3
Spore formation is a form of asexual reproduction used by mushrooms and molds.
During budding, the offspring grows from the body of the parent.
Fragmentation is a form of asexual reproduction that must be followed by regeneration.
Explanation:
Asexual reproduction is the type of reproduction where the gamete formation and fusion have no relevance or existence. It functions on the process of somatic cell division via mitosis and the offsprings are identical to their parents.
The spore formation occurs in fungi through sporangia, bursting open to shed spores, forming into a new young ones. Budding occurs out as an outgrowth of the parent and attains maturity and separates. Fragmentation is the process where the parents fall apart into pieces and regeneration follows.
The second volume : 42.2 L
<h3>Further explanation</h3>
Given
51.7 L at 27 C and 90.9 KPa
Required
The second volume
Solution
STP = P₂=1 atm, T₂=273 K
T₁ = 27 + 273 = 300 K
P₁ = 90.9 kPa = 0,897 atm
Use combine gas law :
P₁V₁/T₁ = P₂V₂/T₂
Input the value :
0.897 x 51.7/300 = 1 x V₂/273
V₂= 42.20 L
