Which list describes the path carbon dioxide takes to the nose?

bronchi, alveolar sacs, larynx, pharynx
trachea, bronchi, capillaries, alveoli
capillaries, trachea, bronchi, larynx
alveolar sacs, bronchi, trachea, pharynx

Alveolar sacs, bronchi, trachea, pharynx : 

Alveolar sacs, bronchi, trachea, pharynx

Respiration is a vital metabolic process that involves exchange of gases like carbon dioxide and oxygen necessary for the survival of all organisms. In humans the respiratory system is composed of different organs like nostrils in the nose, nasal passage, larynx, pharynx, trachea, bronchial tubes, alveoli and lungs. The carbon dioxide is the gas which needs to be expelled out from the body in the process of respiration.

The path of the carbon dioxide takes to the nose is as follows:

1. The oxygen carrying the blood passes from the alveoli into the capillaries, the carbon dioxide already present in the capillaries coming from the blood stream of the entire body, enters into the alveolar sacs.

2. The carbon dioxide containing air then passes to the branching network of bronchial tubes.

3. The carbon dioxide then expelled out from the lungs to the trachea and then from larynx to pharynx.

4. The air enters the nasal passage and finally expelled out from the nostrils .

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol