Explanation:
It is given that r = 0.283 nm. As 1 nm = .
Hence, 0.283 nm =
- Formula for coulombic energy is as follows.
where, e = C
=
=
- As 1 eV =
So, 1 J =
Hence, U =
= 8.9 eV
- Also, 1 J =
= kJ/mol
Therefore, U = kJ/mol
=
Answer : The fraction of carbonic acid present in the blood is 5.95%
Explanation :
The mixture consists of carbonic acid ( H₂CO₃) and bicarbonate ion ( HCO₃⁻). This represents a mixture of weak acid and its conjugate which is a buffer.
The pH of a buffer is calculated using Henderson equation which is given below.
We have been given,
pH = 7.5
pKa of carbonic acid = 6.3
Let us plug in the values in Henderson equation to find the ratio Base/Acid.
The total of mole fraction of acid and base is 1. Therefore we have,
But Base = 15.8 x [Acid]. Let us plug in this value in above equation.
[Acid] = 0.0595 x 100 = 5.95 %
The fraction of carbonic acid present in the blood is 5.95%
Though the ratio of any two atom's masses was the same on either scale, it was horribly confusing, so in 1961, a compromise was reached. Instead of using either Hydrogen, or Oxygen as the standard, the isotope of Carbon<span> with 6 </span>protons<span> and 6 neutrons in its nucleus (</span>Carbon-12<span>) was given a mass of exactly 12.</span>
I think the correct answers are X2Y and X3Y, X2Y5 and X3Y5, and X4Y2 and X3Y,
for the following reason:
If you look at the combining masses of X and Y in
each of the two compounds,
The first compound contains 0.25g of X combined with
0.75g of Y
so the ratio (by mass) of X to Y = 1 : 3
The second compound contains 0.33 g of X combined with
0.67 g of Y
so the ratio (by mass) of X to Y = 1 : 2
Now, you suppose to prepare each of these two
compounds, starting with the same fixed mass of element Y ( I will choose 12g
of Y for an easy calculation!)
The first compound will then contain 4g of X and 12g
of Y
The second compound will then contain 6g of X and
12g of Y
<span>The ratio which combined
the masses of X and the fixed mass (12g) of Y
= 4 : 6
<span>or 2 : 3 </span>
So, the ratio of MOLES of X which combined with the
fixed amount of Y in the two compounds is also = 2 : 3 </span>
The two compounds given with the plausible formula must therefore contain
the same ratio.