Carbon dioxide (co2) is readily soluble in water, according to the equation co2 + h2o ↔ h2co3. carbonic acid (h2co3) is a weak acid. respiring cells release co2 into the bloodstream. what will be the effect on the ph of blood as that blood first comes in contact with respiring cells?

Blood is a buffer solution of bicarbonate (HCO₃⁻) and carbon dioxide (CO₂). The Henderson equation which relates the concentration of HCO₃⁻ and CO₂ is given below:

=.

Respiring cell releases CO₂ in blood stream and that CO₂ on reaction with water molecule produces H₂CO₃ which is a weak base and its conjugate base is  HCO₃⁻.

CO₂ + 2H₂O⇄ HCO₃⁻ + H₃O⁺

pH of the buffer solution (the blood) depends only on the ratio of the amount of CO₂ to the amount of HCO³⁻. [So, due to respiration produced CO₂ will get dissolved in water and favours the equilibrium towards forward direction. Then immediately HCO₃⁻ reacts with HCO₃⁻ and starts producing CO₂.]This ratio remains relatively constant because the concentrations HCO3- and CO2 are very large compared to the amount of CO₂ produced to the blood from respiring cells. So, of blood does not change.

in theory, limestone is composed principally of the compound calcium carbonate - caco3, but in reality this compound is deposited in conditions involving a number of other compounds as impurities so it ends up being a mixture of many compounds.

hope this : -)

tony (lordcaos066)

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