Because chloride is more reactive than hydrogen.
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).
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Answer:
Molecular formula = C₄H₆As₆Cu₄O₁₆
Explanation:
Given data:
Empirical formula = C₂H₃As₃Cu₂O₈
Molar mass of compound = 1013 g/mol
Molecular formula = ?
Solution:
Molecular formula = n (empirical formula)
n = molar mass of compound / empirical formula mass
Empirical formula mass of C₂H₃As₃Cu₂O₈ is 506.897 g/mol
by putting values.
n = 1013 / 506.897
n = 2
Molecular formula = n (empirical formula)
Molecular formula = 2 (C₂H₃As₃Cu₂O₈)
Molecular formula = C₄H₆As₆Cu₄O₁₆
Answer:
sp3d
Explanation:
The ground state electronic configuration of tin is written as; [Kr] 5s²4d¹⁰5p². Hybridization is a concept used to explain the combination of orbitals of appropriate energy to produce suitable orbitals that could be used for bonding.
In forming the compound Snf5^ -1, we have to hybridize the following orbitals on tin; 5p, 5d and 6s orbitals. This gives us a set of sp3d hybrid hence the answer.