The steps to be followed while cleaning volumetric glassware are:
1. Remnants from the previous measurements are wiped off with the help of paper towel.
2. The glassware is then soaked overnight in warm soap solution.
3. Then before rinsing with tap water, the glassware are scrubbed with an appropriate brush.
4. After scrubbing, the glassware is rinsed thoroughly with tap water in order to make sure there are no traces of soap solution.
5. The glassware is then rinsed with de-ionized water and finally with the solution that would be used for the volumetric measurement.
Answer:
Moving Across a Period
Moving from left to right across a period, the atomic radius decreases. The nucleus of the atom gains protons moving from left to right, increasing the positive charge of the nucleus and increasing the attractive force of the nucleus upon the electrons
it would be a crystalline solid, because it could be extended in multiple directions.
When HCl is added to metal ions, metal chlorides are produced. In this problem, it is asked whether the given ions precipitate or not when added to HCl. According to the rule, all chlorides except Ag+, Pb 2+, Hg2 2+ are soluble. Hence the ion that would precipitate is only lead (II) ion.
The reducing agent in the reaction 2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) is lithium (Li).
The general reaction is:
2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) (1)
We can write the above reaction in <u>two reactions</u>, one for oxidation and the other for reduction:
Li⁰(s) → Li⁺(aq) + e⁻ (2)
Fe²⁺(aq) + 2e⁻ → Fe⁰(s) (3)
We can see that Li⁰ is oxidizing to Li⁺ (by <u>losing</u> one electron) in the lithium acetate (<em>reaction 2</em>) and that Fe²⁺ in iron(II) acetate is reducing to Fe⁰ (by <u>gaining</u> two <em>electrons</em>) (<em>reaction 3</em>).
We must remember that the reducing agent is the one that will be oxidized by <u>reducing another element</u> and that the oxidizing agent is the one that will be reduced by <u>oxidizing another species</u>.
In reaction (1), the<em> reducing agent</em> is <em>Li</em> (it is oxidizing to Li⁺), and the <em>oxidizing agent </em>is<em> Fe(CH₃COO)₂</em> (it is reducing to Fe⁰).
Therefore, the reducing agent in reaction (1) is lithium (Li).
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