Answer:
HF
H₂S
H₂CO₃
NH₄⁺
Explanation:
<em>Which acid in each of the following pairs has the stronger conjugate base?</em>
According to Bronsted-Lowry acid-base theory, <em>the weaker an acid, the stronger its conjugate acid</em>. Especially for weak acids, pKa gives information about the strength of such acid. <em>The higher the pKa, the weaker the acid.</em>
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- Of the acids HCl or HF, the one with the stronger conjugate base is HF because it is a weak acid.
- Of the acids H₂S or HNO₂, the one with the stronger conjugate base is H₂S because it is a weaker acid. pKa (H₂S) = 7.04 > pKa (HNO₂) = 3.39
- Of the acids H₂CO₃ or HClO₄, the one with the stronger conjugate base is H₂CO₃ because it is a weak acid.
- Of the acids HF or NH₄⁺, the one with the stronger conjugate base is NH₄⁺ because it is a weaker acid. pKa (HF) = 3.17 < pKa (NH₄⁺) = 9.25
Technically molar mass cannot be in grams, it is in grams per mole. and it refers to a specific number of molecules of a substance, therefore substances have different molar masses because the elements have different weights. for example having 10 water molecules would be a lot heavier than having 10 air molecules
Answer:
A. The average of all the data points
Answer:
Molecular formula
Explanation:
Molecular formula in the first place is required to understand which compound we have. We then should refer to the periodic table and find the molecular weight for each atom. Adding individual molecular weights together would yield the molar mass of a compound.
Then, dividing the total molar mass of a specific atom by the molar mass of a compound and converting into percentage will provide us with the percentage of that specific atom.
E. g., calculate the percent composition of water:
- molecular formula is
; - calculate its molar mass: [tex]M = 2M_H + M_O = 2\cdot 1.00784 g/mol + 16.00 g/mol = 18.016 g/mol;
- find the percentage of hydrogen: [tex]\omega_H = \frac{2\cdot 1.00784 g/mol}{18.016 g/mol}\cdot 100 \% = 11.19 %;
- find the percentage of oxygen: [tex]\omega_O = \frac{16.00 g/mol}{18.016 g/mol}\cdot 100 \% = 88.81 %.
