Answer:
<h3>... :-!...................nose...........</h3>
Given the concentration of aniline hydrochloride is 
Aniline hydrochloride is the conjugate acid of aniline a weak base.
pH can be calculated from
anilinium ion the conjugate acid of aniline.
Answer:
it is energy
Explanation:
Energy is the ability to do work
Answer:
The number of molecules is 1.4140*10^24 molecules
Explanation:
To know the number of molecules, we need to determine how many moles of water we have, water has molar mass of 18.015g/mol
This means that one mole of water molecules has a mass of 18.015g.
42.3g * 1 mole H2O/18.015g
= 2.3480 moles H2O
We are using avogadros number to find the number of molecules of water
2.3480 H2O * 6.022*10^ 23moles/ 1mole of H2O
That's 2.3480 multiplied by 6.022*10^23 divided by 1 mole of H2O
Number of molecules = 1.4140 *10^24 molecules
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:
brainly.com/question/24311846
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