Answer:
Explanation:
first, you calculate the amount of O2 in moles:
98.0 ÷ 32 = 3.0625
second, the ratio if O2/C3H8 is 5 so you need to calculate O2 in moles with that:
3.0625 ÷ 5 = 0.6125
third, the amount of CO2 in moles also can be calculate by the ratio of C3H8/CO2 which is 3
0.6125 × 3 = 1.8375
then multiply CO2 in moles by its molar mass which is 44 g/mol
1.8375 × 44 = 80.85g
Answer:
Concentration: 0.185M HX
Ka = 9.836x10⁻⁶
pKa = 5.01
Explanation:
A weak acid, HX, reacts with NaOH as follows:
HX + NaOH → NaX + H2O
<em>Where 1 mole of HX reacts with 1 mole of NaOH</em>
To solve this question we need to find the moles of NaOH at equivalence point (Were moles HX = Moles NaOH).
18.50mL = 0.01850L * (0.20mol / L) = 0.00370 moles NaOH = Moles HX
In 20.0mL = 0.0200L =
0.00370 moles HX / 0.0200L = 0.185M HX
The equilibrium of HX is:
HX(aq) ⇄ H⁺(aq) + X⁻(aq)
And Ka is defined as:
Ka = [H⁺] [X⁻] / [HX]
<em>Where [H⁺] = [X⁻] because comes from the same equilibrium</em>
As pH = 2.87, [H+] = 10^-pH = 1.349x10⁻³M
Replacing:
Ka = [H⁺] [H⁺] / [HX]
Ka = [1.349x10⁻³M]² / [0.185M]
Ka = 9.836x10⁻⁶
pKa = -log Ka
<h3>pKa = 5.01</h3>
Answer:
Enzyme is carbonic anhydrase
Substrate is 
Turnover number is 
Explanation:
An enzyme is used by a living organism as a catalyst to perform a specific biochemical reaction.
A substrate is a molecule upon which an enzyme acts.
Turnover number refers to the number of substrate molecules transformed by a single enzyme molecule per minute. Here, the enzyme is the rate-limiting factor.
Here,
Enzyme is carbonic anhydrase
Substrate is
Turnover number is
Answer:
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Explanation:
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