Explanation:
Elements need a total of eight electrons to gain stability and look like a noble gas. So, they sometimes need sharing of two, four or even six electrons to complete their octate. So, they form double and triple covalent bonds. One more the reason is the interaction between the p orbitals of the combining atoms. for example A double bond, as in ethene H2C=CH2, arises from one combination of the s orbitals and one combination of the p_y orbitals.
Mass is the inside of an obeject such as what it’s made up of
The concentration required to begin to precipitate PbCl2 for PbCl2 is 0.0216 M.
<h3>
What is molarity?</h3>
Molarity is the measure of the concentration of any solute in per unit volume of the solution.
The reaction is 
The molarity of lead is 0.025 M
The ksp is given 17×10⁻⁵
Now, calculating the concentration
![[Pb^2^+] = 0.025 M.\\Ksp = 1.17 \times 10^-^5\\Ksp = [Pb^2^+] \times [Cl^-]^2\\[Cl^-] = \dfrac{\sqrt{ Ksp}}{[Pb^2^+]} \\\\[Cl^-] = \dfrac{\sqrt{ 0.0000117}}{0.025} \\[Cl^-] = 2.16 \times 10^-^2M.](https://tex.z-dn.net/?f=%5BPb%5E2%5E%2B%5D%20%3D%200.025%20M.%5C%5CKsp%20%3D%201.17%20%5Ctimes%2010%5E-%5E5%5C%5CKsp%20%3D%20%5BPb%5E2%5E%2B%5D%20%5Ctimes%20%20%5BCl%5E-%5D%5E2%5C%5C%5BCl%5E-%5D%20%3D%20%20%5Cdfrac%7B%5Csqrt%7B%20Ksp%7D%7D%7B%5BPb%5E2%5E%2B%5D%7D%20%5C%5C%5C%5C%5BCl%5E-%5D%20%3D%20%20%5Cdfrac%7B%5Csqrt%7B%200.0000117%7D%7D%7B0.025%7D%20%20%5C%5C%5BCl%5E-%5D%20%3D%202.16%20%5Ctimes%2010%5E-%5E2M.)
Thus, the concentration required to begin to precipitate PbCl2 for PbCl2 is 0.0216 M.
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Answer:
1.195 M.
Explanation:
- We can calculate the concentration of the stock solution using the relation:
<em>M = (10Pd)/(molar mass).</em>
Where, M is the molarity of H₂SO₄.
P is the percent of H₂SO₄ (P = 40%).
d is the density of H₂SO₄ (d = 1.17 g/mL).
molar mass of H₂SO₄ = 98 g/mol.
∴ M of stock H₂SO₄ = (10Pd)/(molar mass) = (10)(40%)(1.17 g/mL) / (98 g/mol) = 4.78 M.
- We have the role that the no. of millimoles of a solution before dilution is equal to the no. of millimoles after dilution.
<em>∴ (MV) before dilution = (MV) after dilution</em>
M before dilution = 4.78 M, V before dilution = 250 mL.
M after dilution = ??? M, V after dilution = 1.0 L = 1000 mL.
∴ M after dilution = (MV) before dilution/(V after dilution) = (4.78 M)(250 mL)/(1000 mL) = 1.195 M.
