water molecules are attracted to polar molecules and surround them.
Answer:
A)
Explanation:
It estimates the probability of extinction.
solution:
The quoted atomic mass on the Periodic Table is the WEIGHTED average of the individual isotopic masses. The higher the isotopic percentage, the MORE that isotope will contribute to the isotopic mass. For this reason, most masses that are quoted on the Table are non-integral.
By way of example we could look to the hydrogen atom. The VAST majority of hydrogen atoms (in this universe) are the protium isotope. i.e. 1H, whose nuclei contain JUST the defining proton. There is a smaller percentage (>1%) of hydrogen atoms WITH one NEUTRON in their nuclei to give the deuterium isotope. i.e. 2H, and because this is relatively cheap, and easily incorporated into a molecule, deuterium labelling is routinely used in analysis.
And there is even a smaller percentage of hydrogen atoms with TWO NEUTRONS in their nuclei, to give the tritium isotope. i.e. 3H. The weighted average of the isotopic percentages gives 
It should be for the total solution of 93 plus 20 grams which is 113 grams so 93 divided by 113 grams comes to 82.3% sodium sulfate and this can be checked by multiplying 113 grams by 0.823 which results in 93 grams of sodium sulphate.
Answer:
4.26 %
Explanation:
There is some info missing. I think this is the original question.
<em>Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 × 10
⁻⁴.</em>
<em />
Step 1: Given data
Initial concentration of the acid (Ca): 0.249 M
Acid dissociation constant (Ka): 4.50 × 10
⁻⁴
Step 2: Write the ionization reaction for nitrous acid
HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)
Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])
We will use the following expression.
![[A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4} } = 0.0106 M](https://tex.z-dn.net/?f=%5BA%5E%7B-%7D%20%5D%20%3D%20%5Csqrt%7BCa%20%5Ctimes%20Ka%20%7D%20%3D%20%5Csqrt%7B0.249%20%5Ctimes%204.50%20%5Ctimes%2010%5E%7B-4%7D%20%20%7D%20%3D%200.0106%20M)
Step 4: Calculate the percent ionization of nitrous acid
We will use the following expression.
![\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B%5BA%5E%7B-%7D%20%5D%7D%7B%5BHA%5D%7D%20%5Ctimes%20100%5C%25%20%3D%20%5Cfrac%7B0.0106M%7D%7B0.249%7D%20%5Ctimes%20100%5C%25%20%3D%204.26%5C%25)