Answer:
![m_{PbI_2}=18.2gPbI_2](https://tex.z-dn.net/?f=m_%7BPbI_2%7D%3D18.2gPbI_2)
Explanation:
Hello,
In this case, we write the reaction again:
![Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)](https://tex.z-dn.net/?f=Pb%28NO_3%29_2%28aq%29%20%2B%202%20KI%28aq%29%5Crightarrow%20PbI_2%28s%29%20%2B%202%20KNO_3%28aq%29)
In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:
![n_{Pb(NO_3)_2}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2} *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI} *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{KI}=0.07885molKI](https://tex.z-dn.net/?f=n_%7BPb%28NO_3%29_2%7D%3D%5Cfrac%7B0.14gPb%28NO_3%29_2%7D%7B1g%5C%20sln%7D%2A%5Cfrac%7B1molPb%28NO_3%29_2%7D%7B331.2gPb%28NO_3%29_2%7D%20%20%2A%5Cfrac%7B1.134g%5C%20sln%7D%7B1mL%5C%20sln%7D%20%2A96.7mL%5C%20sln%5C%5C%5C%5Cn_%7BPb%28NO_3%29_2%7D%3D0.04635molPb%28NO_3%29_2%5C%5C%5C%5Cn_%7BKI%7D%3D%5Cfrac%7B0.12gKI%7D%7B1g%5C%20sln%7D%2A%5Cfrac%7B1molKI%7D%7B166.0gKI%7D%20%20%2A%5Cfrac%7B1.093g%5C%20sln%7D%7B1mL%5C%20sln%7D%20%2A99.8mL%5C%20sln%5C%5C%5C%5Cn_%7BKI%7D%3D0.07885molKI)
Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:
![0.04635molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0927molKI](https://tex.z-dn.net/?f=0.04635molPb%28NO_3%29_2%2A%5Cfrac%7B2molKI%7D%7B1molPb%28NO_3%29_2%7D%20%3D0.0927molKI)
But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:
![m_{PbI_2}=0.07885molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2](https://tex.z-dn.net/?f=m_%7BPbI_2%7D%3D0.07885molKI%2A%5Cfrac%7B1molPbI_2%7D%7B2molKI%7D%20%2A%5Cfrac%7B461.01gPbI_2%7D%7B1molPbI_2%7D%20%5C%5C%5C%5Cm_%7BPbI_2%7D%3D18.2gPbI_2)
Best regards.
Answer:
semiconducting and tellurium
Explanation:
u did the test hope this helps babes
360 mg / 1000 => 0.36 g
molar mass => 180 /mol
number of moles:
mass of solute / molar mass
0.36 / 180 => 0.002 moles
Volume solution = 200 mL / 1000 => 0.2 L
M = n / V
M = 0.002 / 0.2
M = 0.01 mol/L
hope this helps!
Answer:
1. earth metal 2. halogen 3.The elements are arranged in seven horizontal rows, called periods or series, and 18 vertical columns, called groups. ... Elements in the periodic table are organized according to their properties.
Inventor: Dmitri Mendeleev please name most brainliest