Answer:
The essence including its given problem is outlined in the following segment on the context..
Explanation:
The given values are:
Moles of CO₂,
x = 0.01962
Moles of water,
Compound's mass,
= 0.4647 g
Let the compound's formula will be:
Combustion's general equation will be:
⇒ 
On putting the estimated values, we get
⇒ 
⇒ 
⇒ 
⇒ 
Now,
x : y : z = 
= 
= 
= 
So that the empirical formula seems to be "C₃H₆O₂".
Answer:
Ciliary body.
Explanation:
Ciliary body: It is the known for the part of the eye that includes the ciliary muscle, which helps in the control the ciliary epithelium and lens shape, which are helping in the production of aqueous humor.
Through active secretion mechanism helping in to produce eighty percent of aqueous humor, and through the plasma ultra-filtration mechanism twenty percent of aqueous humor is produced.
Ciliary body is the part of the layer which helps to deliver the nutrients, and oxygen to the eye tissues, and this layer is known as uvea.
Answer:
B) 0.025 
Explanation:
Solution of the problem is in picture attached,
To solve the problem, we assume the sample to be ideal. Then, we use the ideal gas equation which is expressed as PV = nRT. From the first condition of the nitrogen gas sample, we calculate the number of moles.
n = PV / RT
n = (98.7x 10^3 Pa x 0.01 m^3) / (8.314 Pa m^3/ mol K) x 298.15 K
n = 0.40 mol N2
At the second condition, the number of moles stays the same however pressure and temperature was changed. So, the new volume is calculated as follows:
V = nRT / P
V = 0.40 x 8.314 x 293.15 / 102.7 x 10^3
V = 9.49 x 10^-3 m^3 or 9.49 L
