Answer:
Glycogen. Cellulose. Amylose. Cellulose. Amylopetin and Glycogen. Amylopetin and Cellulose.
Explanation:
Glycogen is the form that glucose is stored in human body.
Cellulose is the structural part of plant cell walls and human cannot digest it.
Amylose is the polysaccharide linked mainly by the the bonds of
1,4 glycosidic.
Cellulose is an unbranched polysaccharide linked mainly by the bonds of
1,4 glycosidic.
Amylopetin and Glycogen are branched polysaccharides linked by the bonds of
1,4 glycosidic and
1,6 glycosidic.
Amylopetin and Cellulose are mainly stored in plants.
Standard Molar Volume is the volume occupied by one mole of any gas at STP. Remember that "STP" is Standard Temperature and Pressure. Standard temperature is 0 &176:C or 273 K. Standard pressure is 1 atmosphere or 760 mm Hg (also called "torr"). 1 mole of any gas at STP occupies 22.4 liters of volume.
The ideal gas law may be written as

where
p = pressure
ρ =density
T = temperature
M = molar mass
R = 8.314 J/(mol-K)
For the given problem,
ρ = 0.09 g/L = 0.09 kg/m³
T = 26°C = 26+273 K = 299 K
M = 1.008 g/mol = 1.008 x 10⁻³ kg/mol
Therefore

Note that 1 atm = 101325 Pa
Therefore
p = 2.2195 x 10⁵ Pa
= 221.95 kPa
= (2.295 x 10⁵)/101325 atm
= 2.19 atm
Answer:
2.2195 x 10⁵ Pa (or 221.95 kPa or 2.19 atm)