Answer
× 10²³ molecules are in 41.8 g of sulfuric acid
Explanation
The first step is to convert 41.8 g of sulfuric acid to moles by dividing the mass of sulfuric acid by its molar mass.
Molar mass of sulfuric acid, H₂SO₄ = 98.079 g/mol
Finally, convert the moles of sulfuric acid to molecules using Avogadro's number.
Conversion factor: 1 mole of any substance = 6.022 × 10²³ molecules.
Therefore, 0.426187053 moles of sulfuric acid is equal
Thus, 2.57 × 10²³ molecules are in 41.8 g of sulfuric acid.
A. The radioactive decay equation is N = N0
where T is the half-life (5730 years), N0 is the number of atoms at time t = 0 and N is the number at time t.
Rewriting this as:
(N/N0) =
Since N = (1/8) N0 and substituting known values:
1/8 =
Taking ln of both sides:
ln(1/8)= -ln(2)*t/5730
t = - 5730 * ln(1/8) / ln (2)
t = 17,190 years
The tree was cut down 17,190 years ago.
B. N0 = 1,500,000 carbon-14 atoms
Since N = (1/8) N0
N = 187,500 carbon atoms left
Answer:- The Ka for the acid is .
Solution:- In general, monoprotic acid could be represented by HA. The dissociation equation for the ionization of HA is written as:
HA(aq)\rightarrow H^+(aq) + A^-(aq)
Now, we make the ice table for this equation as:
HA(aq)\rightarrow H^+(aq) + A^-(aq)
I 0.25 0 0
C -X +X +X
E (0.25 - X) X X
where, I stands for initial concentration, C stands for change in concentration and E stands for equilibrium concentration.
X is the change in concentration and from ice table it's same as the concentration of hydrogen ion that is calculated from given pH.
Where, Ka is the acid ionization constant. Let's plug in the values.
Let's calculate the value of X first using the equation:
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on taking antilog ob above equation we get:
= 0.00195
So, X = 0.001195
Let's plug in this value of X in the equation:-
So, the value of Ka for butyric acid is .