A simple stomach is a chamber lined by large muscular rugae and tiny gastric glands. the monogastric digestive system has ONE simple stomach that secretes acid with a low ph which destroys bacteria and breaks down materials.
this question is semi hard to answer but i hope this helps ^^
Answer:
(a) 106 kPa
(b) 0.0377 mol
(c) 17.8 cm
Explanation:
(a) There are three forces on the piston. Atmospheric pressure pushing down, weight pulling down, and pressure of the gas pushing up.
∑F = ma
PA − mg − PₐA = 0
P = (PₐA + mg) / A
P = Pₐ + (mg / A)
P = 101,300 Pa + (40.0 N) / (π (0.05 m²))
P = 106,393 Pa
P = 106 kPa
(b) Use ideal gas law.
PV = nRT
(106,393 Pa) (π (0.05 m²) (0.11 m)) = n (8.314 Pa m³/mol/K) (20 + 273.15) K
n = 0.0377 mol
(c) Use ideal gas law to find the new volume of the gas.
PV = nRT
(106,393 Pa) (π (0.05 m²) h) = (0.0377 mol) (8.314 Pa m³/mol/K) (200 + 273.15) K
h = 0.178 m
h = 17.8 cm
Answer:
The final pressure of the whole system is 34.80 atm.
Explanation:
Given that,
Volume = 45.0 ml
Volume of first bulb = 77.0 mL
Pressure = 8.89 atm
Volume of second bulb = 250 mL
Pressure = 2.82 atm
Volume of third bulb = 21.0 mL
Pressure = 8.42 atm
We need to calculate the final pressure of the whole system
Using formula of pressure
Where, 
= pressure of first bulb
= pressure of second bulb
= pressure of third bulb
= initial pressure of tube
= Volume of first bulb
=Volume of second bulb
= Volume of third bulb
= Initial volume of tube
Put the value into the formula
Hence, The final pressure of the whole system is 34.80 atm.
