Answer:
The correct answer is c add more HCO3- by adding NaHCO3.
Explanation:
The reaction mentioned in the question is carried out by bicarbonate buffer system of our body to maintain the normal acid base balance.
Now concentration of the reactant (H2C03) is decreased then NaHCO3 should be added which undergo breakdown to release bicarbonate ions(HCO3-).
The released bicarbonate ions then reacts with H+ to form Carbonic acid(H2CO3).Thus homeostasis of H2CO3 is maintained.
Answer:
Mass of H₂O is 3.0g
Explanation:
The reaction equation is given as:
6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂
Parameters that are known:
Mass of CO₂ used = 7.3g
Unknown: mass of water consumed = ?
Solution
To solve this kind of problem, we simply apply some mole concept relationships.
- First, we work from the known to the unknown. From the problem, we have 7.3g of CO₂ that was used. We can find the number of moles from this value using the expression below:
Number of moles of CO₂ = 
- From this number of moles of CO₂, we can use the balanced equation to relate the number of moles of CO₂ to that of H₂O:
6 moles of CO₂ reacted with 6 moles of H₂O(1:1)
- We can then use the mole relationship with mass to find the unknown.
Workings
>>>> Number of moles of CO₂ =?
Molar mass of CO₂ :
Atomic mass of C = 12g
Atomic mass of O = 16g
Molar mass of CO₂ = 12 + (2 x16) = 44gmol⁻¹
Number of moles of CO₂ =
= 0.166moles
>>>>>> if 6 moles of CO₂ reacted with 6 moles of H₂O, then 0.166moles of CO₂ would produce 0.166moles of H₂O
>>>>>> Mass of water consumed = number of mole of H₂O x molar mass
Mass of H₂0 = 0.166 x ?
Molar mass of H₂O:
Atomic mass of H = 1g
Atomic mass of O = 16
Molar mass of H₂O = (2x1) + 16 = 18gmol⁻¹
Mass of H₂O = 0.166 x 18 = 3.0g
<span>The half-life of Carbon 14 and radionuclides are used to estimate the absolute (versus relative) age of pre-history items </span>
The closer you get to the equator the warmest you will be. 10 degrees north is warmer than 80 degrees north, because 10 degrees north is closer to the equator. I don't know if that was helpful or not but, I tried to answer as best as I could.