The answeris c, chemicalchange because the atoms in woodand oxygen are rearranged.
Answer:
Explanation:
A buffer is defined as an aqueous mixture of a weak acid and its conjugate base or vice versa.
In the systems:
H₂CO₃(aq) and KHCO₃(aq): Carbonic acid, H₂CO₃, is a weak acid that, in solution with its conjugate pair, HCO₃⁻ make a <em>buffer system.</em>
NaCl(aq) and NaOH(aq): NaCl is a salt and NaOH is a strong base. Thus, this system <em>is not </em> a buffer system.
H₂O(l) and HCl(aq): Water is a solvent and HCl a strong acid. This <em>is not </em>a buffer system.
HCl(aq) and NaOH(aq): HCl is a strong acid and NaOH a strong base. This <em>is not </em>a buffer system.
NaCl(aq) and NaNO₃(aq): Both NaCl and NaNO₃ are salts and this system <em>is not </em>a buffer system.
The question is incomplete, here is the complete question:
What volume (mL) of the partially neutralized stomach acid having concentration 2 M was neutralized by 0.1 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)
<u>Answer:</u> The volume of HCl neutralized is 1.25 mL
<u>Explanation:</u>
To calculate the volume of acid, we use the equation given by neutralization reaction:

where,
are the n-factor, molarity and volume of stomach acid which is HCl
are the n-factor, molarity and volume of base which is NaOH.
We are given:

Putting values in above equation, we get:

Hence, the volume of HCl neutralized is 1.25 mL
Impure substance, because pure would be dirt from the earth
Given the temperature 746 K and activity of Pb equal to 0.055. The mole fraction of Pb is 0.1. So, the mole fraction of Sn = 0.9.Activity coefficient, γ = 0.055 / 0.1 = 0.55.The expression for w=ln〖γ_Pb x RT〗/(X_Sn^2 )=(-0.5978 x 8.314 J/(mol K ) x 746 K)/(0.9 x 0.9)= -4577.7 J= -4578 J
Now we use the computed value above and new temperature 773 K. The mole fraction of Sn and Pb are 0.5 and 0.5 respectively. Calculate the activity coefficient in the following manner.lnγ_Sn=w/RT X_Pb^2=(-4578 J)/(8.314 J/mol x 773 K) x 0.5 x 0.5= -0.718lnγ_Sn=exp(-0.178)=0.386The activity of Sn= γ_Sn x X_Sn=0.386 x 0.5=0.418
w of the system is -4578 J and the activity of Sn in the liquid solution of xsn at 500 degree Celsius is 0.418