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3 years ago

The atomic mall mass is an element depends upon the.

2 answers:

8 0

The atomic mass of element has been dependent on mass of each isotopes and their abundance in nature.

The atomic mass has been the mass of neutrons and protons of the element. There has been the presence of the isotopes of the elements, with varying abundance, and varying atomic masses.

The atomic mass of the element has been the relative average atomic mass of the element with respect to the mass of the isotopes and their percent abundance.

The average atomic mass (amu) has been given by:

$amu=M_1A_1\;+\;M_2A_2$

Where, $M_1A_1$ has been the atomic mass and abundance of isotope 1.

$M_2A_2$ has been the atomic mass and abundance of isotope 2.

Thus, the average atomic mass has been dependent on mass of each isotopes and their abundance in nature.

For more information about the atomic mass, refer to the link:

brainly.com/question/5566317

evablogger [386]3 years ago

7 0

Answer:

Mass and relative abundance of each isotope of that element

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A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb

andre [41]

Answer:

$m_{PbI_2}=18.2gPbI_2$

Explanation:

Hello,

In this case, we write the reaction again:

$Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)$

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:

$n_{Pb(NO_3)_2}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2} *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI} *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{KI}=0.07885molKI$

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:

$0.04635molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0927molKI$

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:

$m_{PbI_2}=0.07885molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2$

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