Answer:
(C3H4O3) x 2 = C6H8O6, the molecular formula for Vitamin C.
Since 1 mole of any element contains 6.0221023 atoms, a mole of carbon and a mole of oxygen will have the same number of atoms.
<h3>Is a mole of oxygen equivalent to a mole of carbon?</h3>
Answer and justification The molar mass of carbon differs from the molar mass of oxygen, hence this assertion is untrue. As a result, one mole of oxygen and one mole of carbon cannot have the same mass.
<h3>What kind of link unites carbon and oxygen?</h3>
A polar covalent link between carbon and oxygen is known as a carbon-oxygen bond. With six valence electrons, oxygen likes to either share two of them in a bond with carbon, leaving the remaining four unoccupied.
To know more about mole of carbon atoms visit:-
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The pH of a buffer solution : 4.3
<h3>Further explanation</h3>
Given
0.2 mole HCNO
0.8 mole NaCNO
1 L solution
Required
pH buffer
Solution
Acid buffer solutions consist of weak acids HCNO and their salts NaCNO.
![\tt \displaystyle [H^+]=Ka\times\frac{mole\:weak\:acid}{mole\:salt\times valence}](https://tex.z-dn.net/?f=%5Ctt%20%5Cdisplaystyle%20%5BH%5E%2B%5D%3DKa%5Ctimes%5Cfrac%7Bmole%5C%3Aweak%5C%3Aacid%7D%7Bmole%5C%3Asalt%5Ctimes%20valence%7D)
valence according to the amount of salt anion
Input the value :
![\tt \displaystyle [H^+]=2.10^{-4}\times\frac{0.2}{0.8\times 1}\\\\(H^+]=5\times 10^{-5}\\\\pH=5-log~5\\\\pH=4.3](https://tex.z-dn.net/?f=%5Ctt%20%5Cdisplaystyle%20%5BH%5E%2B%5D%3D2.10%5E%7B-4%7D%5Ctimes%5Cfrac%7B0.2%7D%7B0.8%5Ctimes%201%7D%5C%5C%5C%5C%28H%5E%2B%5D%3D5%5Ctimes%2010%5E%7B-5%7D%5C%5C%5C%5CpH%3D5-log~5%5C%5C%5C%5CpH%3D4.3)
NaOH +HCl ---> NaCl +H2O
n (mol HCl) = n (mol NaOH)
M- molarity
V - volume
M(HCl)V(HCl) = M(NaOH)V(NaOH)
M(HCl)= M(NaOH)V(NaOH)/V(HCl)
M(HCl)= 10⁻³*25.3 ml/5.25 ml=<span>4.82 × 10⁻³ M molarity HCl</span>
Answer:
0.8988
Explanation:
To calculate the fractional saturation of hemoglobin , the formula used is
YO_2=![\frac{(pO_2)^n}{(p50)^n+(pO_2)^n}](https://tex.z-dn.net/?f=%5Cfrac%7B%28pO_2%29%5En%7D%7B%28p50%29%5En%2B%28pO_2%29%5En%7D)
now putting the values
YO_2= ![\frac{58^3}{28^3+58^3}](https://tex.z-dn.net/?f=%5Cfrac%7B58%5E3%7D%7B28%5E3%2B58%5E3%7D)
= ![\frac{195112}{21952+195112}](https://tex.z-dn.net/?f=%5Cfrac%7B195112%7D%7B21952%2B195112%7D)
=0.8988
Therefore, fractional saturation of hemoglobin= 0.8988
where
YO_2= fractional saturation of hemoglobin
pO_2= partial pressure of oxygen
p50= is the pO_2 at which hemoglobin is 50% saturated