We are given with a 2.5 M stock solution of acetic acid and we are required to calculate the volume of the solution needed to prepare 100 milliliters of 0.5 M acetic acid solution. To solve this, we acquire the formula <span>Mconcentrated*Vconcentrated = Mdilute*Vdilute. That is 2.5 M*x=0.5M*100 ml where x is the volume of 2.5 M needed. x is equal to 20 ml. So we need 20 ml of 2.5 M solution and dilute to 100 ml using water as diluent.</span>
For multiple covalent bonds to form in molecules, the molecules must contain carbon nitrogen or oxygen.
<u>Explanation:</u>
- Think about carbon dioxide (CO2). If every oxygen atom imparts one electron to the carbon molecule, there will be 6 electrons in carbon particle and 7 electrons in every oxygen atom. This doesn't give the carbon atom as a total octet.
- Sometimes more than one set of electrons is shared between two atoms. In carbon dioxide, a second electron from every oxygen atom is likewise imparted with the central carbon atom, and the carbon particle imparts one more electron with every oxygen atom.
- Two sets of electrons shared between two atoms make a double bond between the atoms. A few particles contain triple bonds, covalent bonds in which three sets of electrons are shared by two atoms.
Answer : The percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.
Explanation :
First we have to calculate the number of atoms in 5.5 wt% Pb and 94.5 wt% of Sn.
As, 207.2 g of lead contains
atoms
So, 5.5 g of lead contains
atoms
and,
As, 118.71 g of lead contains
atoms
So, 94.5 g of lead contains
atoms
Now we have to calculate the percent composition of Pb and Sn in atom.
![\% \text{Composition of Pb}=\frac{\text{Atoms of Pb}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100](https://tex.z-dn.net/?f=%5C%25%20%5Ctext%7BComposition%20of%20Pb%7D%3D%5Cfrac%7B%5Ctext%7BAtoms%20of%20Pb%7D%7D%7B%5Ctext%7BAtoms%20of%20Pb%7D%2B%5Ctext%7BAtoms%20of%20Sn%7D%7D%5Ctimes%20100)
![\% \text{Composition of Pb}=\frac{1.59\times 10^{22}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=3.21\%](https://tex.z-dn.net/?f=%5C%25%20%5Ctext%7BComposition%20of%20Pb%7D%3D%5Cfrac%7B1.59%5Ctimes%2010%5E%7B22%7D%7D%7B%281.59%5Ctimes%2010%5E%7B22%7D%29%2B%284.79%5Ctimes%2010%5E%7B23%7D%29%7D%5Ctimes%20100%3D3.21%5C%25)
and,
![\% \text{Composition of Sn}=\frac{\text{Atoms of Sn}}{\text{Atoms of Pb}+\text{Atoms of Sn}}\times 100](https://tex.z-dn.net/?f=%5C%25%20%5Ctext%7BComposition%20of%20Sn%7D%3D%5Cfrac%7B%5Ctext%7BAtoms%20of%20Sn%7D%7D%7B%5Ctext%7BAtoms%20of%20Pb%7D%2B%5Ctext%7BAtoms%20of%20Sn%7D%7D%5Ctimes%20100)
![\% \text{Composition of Sn}=\frac{4.79\times 10^{23}}{(1.59\times 10^{22})+(4.79\times 10^{23})}\times 100=96.8\%](https://tex.z-dn.net/?f=%5C%25%20%5Ctext%7BComposition%20of%20Sn%7D%3D%5Cfrac%7B4.79%5Ctimes%2010%5E%7B23%7D%7D%7B%281.59%5Ctimes%2010%5E%7B22%7D%29%2B%284.79%5Ctimes%2010%5E%7B23%7D%29%7D%5Ctimes%20100%3D96.8%5C%25)
Thus, the percent composition of Pb and Sn in atom is, 3.21 % and 96.8 % respectively.
Explanation:
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