H2SO3 or sulfurous acid is actually a strong acid. We know
for a fact that strong acids completely dissociate into its component ions in a
solution, that is:
<span>H2SO3 --> 2H+ + SO3-</span>
<span>So from the equation above, there are 2 moles of H+</span>
Answer:
75 mg
Explanation:
We can write the extraction formula as
x = m/[1 + (1/K)(Vaq/Vo)], where
x = mass extracted
m = total mass of solute
K = distribution coefficient
Vo = volume of organic layer
Vaq = volume of aqueous layer
Data:
m = 75 mg
K = 1.8
Vo = 0.90 mL
Vaq = 1.00 mL
Calculations:
For each extraction,
1 + (1/K)(Vaq/Vo) = 1 + (1/1.8)(1.00/0.90) = 1 + 0.62 = 1.62
x = m/1.62 = 0.618m
So, 61.8 % of the solute is extracted in each step.
In other words, 38.2 % of the solute remains.
Let r = the amount remaining after n extractions. Then
r = m(0.382)^n.
If n = 7,
r = 75(0.382)^7 = 75 × 0.001 18 = 0.088 mg
m = 75 - 0.088 = 75 mg
After seven extractions, 75 mg (99.999 %) of the solute will be extracted.
The alcohol concentration of the mixed solution is 20%
Simplification :
Based on the given condition, formulate :
35% ×0.40 + 0.6 ×10% ÷{ 0.4+0.6}
Calculate the product :
Calculate the sum or difference : 
Any fraction with denominator 1 is equal to numerator : 0.2
Multiply a number to both numerator, denominator : 0.2 ×
Calculate the product or quotient : 
A fraction with denominator equals to 100 to a percentage 20%.
How do you find the concentration of a mixed solution?
In general when your are mixing two different concentrations together first calculate number of moles for each solution (n=CV ,V-in liter) then add them together it will be total moles,then concentration of mixture will be = total moles / total volume(liter).
Learn more about concentration of alcohol :
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<u>Answer:</u> The percentage abundance of
and
isotopes are 77.5% and 22.5% respectively.
<u>Explanation:</u>
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
.....(1)
Let the fractional abundance of
isotope be 'x'. So, fractional abundance of
isotope will be '1 - x'
- <u>For
isotope:</u>
Mass of
isotope = 35 amu
Fractional abundance of
isotope = x
- <u>For
isotope:</u>
Mass of
isotope = 37 amu
Fractional abundance of
isotope = 1 - x
Average atomic mass of chlorine = 35.45 amu
Putting values in equation 1, we get:
![35.45=[(35\times x)+(37\times (1-x))]\\\\x=0.775](https://tex.z-dn.net/?f=35.45%3D%5B%2835%5Ctimes%20x%29%2B%2837%5Ctimes%20%281-x%29%29%5D%5C%5C%5C%5Cx%3D0.775)
Percentage abundance of
isotope = 
Percentage abundance of
isotope = 
Hence, the percentage abundance of
and
isotopes are 77.5% and 22.5% respectively.
Answer:
hope its help you
Explanation:
2.600 × 10^-5 × 6.100 × 10^-5 = 1.586 × 10-9