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
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If the reaction is represented by:
PCl₃ + Cl₂ <-> PCl₅ (exothermic)
the mole fraction of chlorine in the equilibrium mixture will change according to the following:
Decrease the volume: decrease
Increase the temperature: increase
Increase the volume: increase
Decrease the temperature: decrease
Answer: B. No, fluorine is always assigned an oxidation number of -1.
Explanation:
Answer:
357 g of the transition metal are present in 630 grams of the compound of the transition metal and iodine
Explanation:
In any sample of the compound, the percentage by mass of the transition metal is 56.7%. This means that for a 100 g sample of the compound, 56.7 g is the metal while the remaining mass, 43.3 g is iodine.
Given mass of sample compound = 630 g
Calculating the mass of iodine present involves multiplying the percentage by mass composition of the metal by the mass of the given sample;
56.7 % = 56.7/100 = 0.567
Mass of transition metal = 0.567 * 630 = 357.21 g
Therefore, the mass of the transition metal present in 630 g of the compound is approximately 357 g
Not sure about any others but I believe 3 goes with B
