By definition, Bronsted-Lowry acid is a proton donor in the acid-base neutralization reaction. When a weak acid like acetylsalicylic acid is reacted with water, the water here acts as the Bronsted-Lowry base. This is possible because water has properties of amphoterism - can act as an acid or base. In this case, acetylsalicylic acid would have to donate its H+ atom to water, so that it would yield a hydronium ion, H₃O⁺. The complete net ionic reaction is shown in the picture.
So, in the reaction, the products yield are the acetylsalicylate ion and the hydronium ion.
oops pls forgive me I accidentally did the wrong question.
Pure air is a mixture of several gases that are invisible and odorless. Consists of 78% nitrogen, 21% oxygen, and less than 1% of argon, carbon dioxide.
The isotope that is more abundant, given the data is isotope Li7
<h3>Assumption</h3>
- Let Li6 be isotope A
- Let Li7 be isotope B
<h3>How to determine whiche isotope is more abundant</h3>
- Molar mass of isotope A (Li6) = 6.02 u
- Molar mass of isotope B (Li7) = 7.02 u
- Atomic mass of lithium = 6.94 u
- Abundance of A = A%
- Abundance of B = (100 - A)%
Atomic mass = [(mass of A × A%) / 100] + [(mass of B × B%) / 100]
6.94 = [(6.02 × A%) / 100] + [(7.02 × (100 - A)) / 100]
6.94 = [6.02A% / 100] + [702 - 7.02A% / 100]
6.94 = [6.02A% + 702 - 7.02A%] / 100
Cross multiply
6.02A% + 702 - 7.02A% = 6.94 × 100
6.02A% + 702 - 7.02A% = 694
Collect like terms
6.02A% - 7.02A% = 694 - 702
-A% = -8
A% = 8%
Thus,
Abundance of B = (100 - A)%
Abundance of B = (100 - 8)%
Abundance of B = 92%
SUMMARY
- Abundance of A (Li6) = 8%
- Abundance of B (Li7) = 92%
From the above, isotope Li7 is more abundant.
Learn more about isotope:
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Answer:
12.09 L
Explanation:
Step 1: Convert 826.1 mmHg to atm
We will use the conversion factor 760 mmHg = 1 atm.
826.1 mmHg × 1 atm/760 mmHg = 1.087 atm
Step 2: Convert 427.8 J to L.atm
We will use the conversion factor 101.3 J = 1 L.atm.
427.8 J × 1 L.atm/101.3 J = 4.223 L.atm
Step 3: Calculate the change in the volume
Assuming the work done (w) is 4.223 L.atm against a pressure (P) of 1.087 atm, the change in the volume is:
w = P × ΔV
ΔV = w/P
ΔV = 4.223 L.atm/1.087 atm = 3.885 L
Step 4: Calculate the final volume
V₂ = V₁ + ΔV
V₂ = 8.20 L + 3.885 L = 12.09 L
