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3 years ago

A transition metal in the fourth period from the following list : Cu, O , Pr, Ag

Chemistry

1 answer:

Kaylis [27]3 years ago

3 0

Answer:

Explanation:

Groups 3 - 12 (or groups IIA - IIB) of the periodic table contain transition elements. Transaction elements start from period four (4) of the periodic table. The phrase alludes to the fact that the d sublevel is filling at a lower main energy level than the s sublevel that came before it.

The transition elements' arrangement is inverted from the fill order, with the 4 s filled prior to the actual 3 d begins. The transition elements are commonly referred to as transition metals since they are all metals. They are less reactive than the metals in Groups 1 and 2 and have normal metallic characteristics.

From the options given Cu is the only transition metal in the fourth period on the periodic table.

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3 years ago

Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3)2(aq)+2

Setler79 [48]

Answer:

a) volume of ammonium iodide required =349 mL

b) the moles of lead iodide formed = 0.0436 mol

Explanation:

The reaction is:

$Pb(NO_{3})_{2}+2NH_{4}I -->PbI_{2}+2NH_{4}NO_{3}$

It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.

Let us calculate the moles of lead nitrate taken in the solution.

Moles=molarityX volume (L)

Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol

the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol

The volume of ammonium iodide required will be:

$volume=\frac{moles}{molarity}=\frac{0.0872}{0.250}=0.349L=349mL$

the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol

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