Freediving is an activity in which a person dives, sometimes to great depth, without the use of scuba gear. the diver must hold his or her breath for the duration of the dive. the record depth, with no equipment such as diving fins, is 101 m; the official record time is over 11 minutes. historically, freediving has been used by pearl divers and sponge divers. some freedivers hyperventilate (breathe rapidly and deeply) before diving. hyperventilation can change the concentration of co2 in the blood and may increase the length of time that a person feels like he or she can hold his or her breath. how does hyperventilation affect blood ph?

Explanation:

Carbon dioxide is eliminated from the body as a gas through exhaled air. When the volumes of air breathed into and out of the lungs increases above what is normal (hyperventilation), more carbon dioxide than normal is eliminated.

When carbon dioxide is dissolved in water forms carbonic acid which is an acid, having less CO2 in the system affects the balance of the normal pH tips towards the blood pH to become alkaline (higher pH), this condition is called respiratory alkalosis.

We can observe the equilibrium in this equation:

CO2 + H2O ← H2CO3 ← H+ + HCO3

Here we observe how losing CO2 during hyperventilation leads to a decrease of H+ and therefore an increase in pH.

I hope you find this information useful and interesting! Good luck!

the answer is the first option

explanation:

just googled it m8. not that hard

lol

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