6.02 × 10^23 hydrogen ions per mol
Answer:
[HAc] = 0.05M
[Ac⁻] = 0.20M
Explanation:
The Henderson-Hasselbalch formula for the acetic acid buffer is:
pH = pka + log₁₀ [Ac⁻] / [HAc]
Replacing:
5.36 = 4.76 + log₁₀ [Ac⁻] / [HAc]
3.981 = [Ac⁻] / [HAc] <em>(1)</em>
Also, as total concentration of buffer is 0.25M it is possible to write:
0.25M = [Ac⁻] + [HAc] <em>(2)</em>
Replacing (2) in (1)
3.981 = 0.25M - [HAc] / [HAc]
3.981 [HAc] = 0.25M - [HAc]
4.981 [HAc] = 0.25M
<em>[HAc] = 0.05M</em>
Replacing this value in (2):
0.25M = [Ac⁻] + 0.05M
<em>[Ac⁻] = 0.20M</em>
I hope it helps!
For a (unbalanced) reaction: NaOH +CO2-Na2CO3 + H2O, the moles of NaOH and moles of each product are formed are mathematically given as
a) Moles of NaOH =44.05
b) Moles of Na2CO3=21.0
<h3>What is the moles of NaOH and what moles of each product are formed?</h3>
Generally, the equation for the Chemical reaction is mathematically given as
2 NaOH(aq)+ CO2(g)------> Na2CO3(aq)+ H2O(l)
Therefore
Moles of CO2= 925/44
Moles of CO2=21.0
Hence
Moles of NaOH = 2 x Moles of CO2
Moles of NaOH = 2x925/44
Moles of NaOH =44.05
In conclusion
Moles of Na2CO3 925/44
Moles of Na2CO3=21.0
And
Moles of H2O= 925/44
Moles of H2O= 21.0
Read more about Chemical reaction
brainly.com/question/16416932
The correct answer is option B. i.e. the number of protons is increased.
In beta-particle emission, a neutron is changed into proton, electron and electron neutrino. The electron is emitted as beta particle and the proton remains inside the atom. Thus, the mass number remains same but the atomic number increases by one.
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