Answer:
The H+ (aq) concentration of the resulting solution is 4.1 mol/dm³
(Option C)
Explanation:
Given;
concentration of HA,
= 6.0mol/dm³
volume of HA,
= 25.0cm³, = 0.025dm³
Concentration of HB,
= 3.0mol/dm³
volume of HB,
= 45.0cm³ = 0.045dm³
To determine the H+ (aq) concentration in mol/dm³ in the resulting solution, we apply concentration formula;

where;
is initial concentration
is initial volume
is final concentration of the solution
is final volume of the solution

Therefore, the H+ (aq) concentration of the resulting solution is 4.1 mol/dm³
The first step is to calculate the molarity of each compound:
final volume of solution = 157 + 139 = 296 mL
molarity of <span>nac2h3o2 = (157 x 0.35) / 296 = 0.1856 molar
molarity of </span><span>hc2h3o2 = (139 x 0.46) / 296 = 0.216 molar
Then, we calculate the pH as follows:
pKa of acetic acid = -log(</span><span>1.75 × 10^-5) = 4.7569
pH = pKa + </span><span> log ([salt] / [acid])
= </span>4.7569 + log(0.1856 / 0.216)
= 4.691
Sedimentary rock can be minazute by adding more heat n pressure
Answer : The correct option is, the negative log of the hydroxide ion concentration.
Explanation :
pOH : It is defined as the negative logarithm of hydroxide ion concentration. It is a measure of the alkalinity of the solution.
Formula used :
![pOH=-log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-log%5BOH%5E-%5D)
is the concentration of
ions.
When pOH is less than 7, the solution is alkaline.
When pOH is more than 7, the solution is acidic.
When pOH is equal to 7, the solution is neutral.
Answer:
The answer to your question is V2 = 825.5 ml
Explanation:
Data
Volume 1 = 750 ml
Temperature 1 = 25°C
Volume 2= ?
Temperature 2 = 55°C
Process
Use the Charles' law to solve this problem
V1/T1 = V2/T2
-Solve for V2
V2 = V1T2 / T1
-Convert temperature to °K
T1 = 25 + 273 = 298°K
T2 = 55 + 273 = 328°K
-Substitution
V2 = (750 x 328) / 298
-Simplification
V2 = 246000 / 298
-Result
V2 = 825.5 ml