A 250 ml sample of saturated a g o h solution was titrated with h c l , and the endpoint was reached after 2. 60 ml of 0. 0136 m h c l was dispensed. Based on this titration, what is the k s p of a g o h <u>. Ksp=1.9×10⁻⁸</u>
<h3>What is titration?</h3>
Titration is a typical laboratory technique for quantitative chemical analysis used to calculate the concentration of a specified analyte. It is also referred to as titrimetry and volumetric analysis (a substance to be analyzed). A standard solution with a known concentration and volume is prepared as the reagent, also known as the titrant or titrator. To ascertain the concentration of the analyte, the titrant reacts with an analyte solution (also known as the titrand). The titration volume is the amount of titrant that interacted with the analyte.
A typical titration starts with a beaker or Erlenmeyer flask being placed below a calibrated burette or chemical pipetting syringe that contains the titrant and a little amount of the indicator (such as phenolphthalein).
To learn more about titration from the given link:
brainly.com/question/186765
#SPJ4
The common substance among the product(s) of the first equation and among the reactant(s) in the second equation is H2O(g). We can eliminate that as an intermediate. The overall chemical equation will thus be:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l),
which is the first answer choice.
In essence, all you’re doing here is swapping water vapor for liquid water.
Answer: 0.52V
Explanation:
Ecell = Ecell(standard) - [(0.0592 logQ)/n]
Q = product of the quotient
n = no of electrons transferred = 2
Ecell = 0.63 - [(0.0592*Log(1 / 2.0 * 10-4) / 2]
Ecell = 0.63 - 0.0194
Ecell = 0.5205V
methanol:
1 mole CH3 OH --> produces --> 1 mole CO2
1 mole CO2 has a molar mass of 44.01 gh/mole
your set up is:
(44.01 g CO2) / -726.5kJ = 0.06058g
your answer 0.06058 grams of CO2 produced per kJ released.
