A chemistry student is performing a lab exercise where they are reacting aqueous Lead (II) Nitrate with aqueous Potassium Iodide. The student begins the reaction with 10.0 g of Lead (II) Nitrate and 12.0 g of Potassium Iodide. At the end of the experiment the student has collected 11.4 grams of the bright yellow solid precipitate Lead (II) Iodide. Write the full balanced equation. Include the appropriate states of matter.
How many atoms of Lead are found in the product collected by the student?
Based upon the information above, what is the percent yield of this student’s lab?

 Pb(NO3)2(aq) + 2KI(aq)  —> PbI2(s) + 2KNO3(aq)

1.57x10^22 atoms of Pb

87% yield

Explanation:

10.0 g Pb(II) nitrate = 10/333 moles = 0.03 moles

12.0 g KI = 12/166 moles = 0.072 moles (0.06 moles required)

11.4 g PbI2 = 11.4/421 moles = 0.026 moles

expected yield 0.03 moles, yield = 87%

atoms in 11.4 g = 0.026 moles = 0.026*6.02214076*10^23 atoms = 1.57*10^22

answer: the figure is attached herewith.

step-by-step explanation:   we are given to sketch a unit circle and label the terminal points corresponding to

in the attached figure, we have drawn a unit circle, with centre at o(0,0) and radius 1 unit. we have labelled the points a, b, c, d and e corresponding to the angles below

∠aox = 60°,

∠box = 120°,

∠cox = 180°,

∠dox = 240°,

∠eox = 300°.

and the coordinates are a(0.5,1), b(-0.5,1), c(-1,0), d(-0.5,-1) and e(0.5,-1).

thus, the answer is complete.

cytokinesis is the splitting of a cell membrane to form 2 identical daughter cells. it's similar to mitosis as both cells have identical copies of dna and cytokinesis is a step in the mitotic cycle.