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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<span>osmotic pressure = CRT
to get C, get the number of moles of NaCl then divide it with volume of the solution
mole NaCl = 0.923g / 58.442 g/mol
mole NaCl = 0.0158 mol
C = 0.0158 mol / 0.1L
C = 0.158 mol/L
So,
osmotic pressure = 0.158 mol/L*(0.082058 L-atm/mol-K)*(25+273K)
osmotic pressure = 3.86 atm</span>
Answer:
1. grasslands
Explanation: hope this helps :)
Answer:
kindly refer to the picture attached
Explanation:
Have a great day . keep smiling
Answer:
C₂H₄O
Explanation:
In a compound that contains Cabon, hydrogen and oxygen, the combustion produce CO₂ from the carbon, and H₂O from the hydrogens. Using the mass of the products we can solve the moles of Carbon and hydrogen. The empirical formula is the simplest whole-number of atoms present in a molecule.
<em>Moles CO₂ = Moles C:</em>
11.8g CO₂ * (1mol / 44g) = 0.268 moles CO₂ = 0.268 moles C * (12g/mol) =
3.216g C
<em>Moles H₂O = 1/2 moles H:</em>
4.83g H₂O * (1mol / 18g) = 0.268 moles H₂O * (2 mol H / 1 mol H₂O) =
0.537 mol H * (1g/mol) = 0.537g H
<em>Mass O to find moles O:</em>
5.90g Sample - 3.216g C - 0.537g H = 2.147g O * (1mol / 16g) = 0.134 moles O
<em>Ratio of atoms -Dividing in 0.134 moles-:</em>
C = 0.268mol C / 0.134 mol O = 2
H = 0.537mol H / 0.134 mol O = 4
O = 0.134mol O / 0.134 mol O = 1
Empirical formula is:
<h3>C₂H₄O</h3>