Answer:

Explanation:
You look at the type of atom and their electronegativity difference.
If ΔEN <1.6, covalent; if ΔEN >1.6, ionic
Ar/Xe: Noble gases; no reaction
F/Cs: Non-metal + metal; ΔEN = |3.98 – 0.79| = 3.19; Ionic
N/Br: Two nonmetals; ΔEN = |3.04 - 2.98| = 0.
Let MM(x) be the molar mass of x.
MM(Pb) : MM(PbO)
=207.21 : 223.20 = 451.4 g : x g
cross multiply and solve for x
x=223.2/207.21*451.4
= 486.23 g
Percentage yield = 365.0/486.23= 0.75067 = 75.07% (rounded to 4 sign. fig.)
Elements were grouped on their properties and behaviors, so hydrogen resides with the alkali metals in group 1 (1A) because it has only 1 valence electron, like the metals in that group.
Oxidation, Dissolving in acid, Hydrolysis ,Water absorption and Hydration
Taking the average of more measurements decreases random error of measurement
Taking the average of many measurements is the most effective way to reduce random errors in a measurement. Because the certainty of the results grows as the number of data does, Less risk of random errors means that the value is more certain. Fewer measurements lead to less reliable data collection, which raises the likelihood of random errors.
The complete question is
Which procedure(s) decrease(s) the random error of a measurement: (1) taking the average of more measurements: (2) calibrating the instrument; (3) taking fewer measurements? Explain
To learn more about random errors:
brainly.com/question/14149934
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