The answer is likely to be A
Answer:
The answer to your question is 0.269 g of Pb
Explanation:
Data
Lead solution = 0.000013 M
Volume = 100 L
mass = 0.269 g
atomic mass Pb = 207.2 g
Chemical reaction
2Pb(s) + O₂(aq) + 4H⁺(aq) → 2H₂O(l) + 2Pb₂⁺(aq)
Process
1.- Calculate the mass of Pb in solution
Formula
Molarity = 
Solve for number of moles
Number of moles = Volume x Molarity
Substitution
Number of moles = 100 x 0.000013
Number of moles = 0.0013
2.- Calculate the mass of Pb formed.
207.2 g of Pb ----------------- 1 mol
x g ----------------- 0.0013 moles
x = (0.0013 x 207.2) / 1
x = 0.269 g of Pb
The gibbs free energy of the reaction of diamond to graphite is equal to -2.90 kJ/mol. The free energy is negative which means that the reaction is spontaneous. Therefore, the forward reaction is favored. Hope this helps. Have a nice day.
Answer:
Elements form compounds to satisfy the octet rule. Noble gasses never form compounds because they already satisfy the octet rule.
Explanation:
The octet Rule is the theory that an element will attempt to gain a valence of 8 by binding with another element in it's vicinity. This can happen in a variety of ways, but the main thing to remember is that they will take the "shortest path" to 8(I.e an element will sometimes lose an electron or 2 if it has a valence 1 or 2 to loop back around to 8, while an element with a valence of 6 or 7 will attempt to gain 2 or 1 electrons).
Valence of elements can be counted by group in the image attached.
Group 1 has a valence of 1, Group 2 has a valence of 2, then we move to group 13 which has a valence of 3, group 14 has a valence of 4, group 15 has a valence of 5, group 16 has 6, group 17 has 7, and group 18 is the noble gasses which have 8.
Answer:
Solubility of O₂(g) in 4L water = 3.42 x 10⁻² grams O₂(g)
Explanation:
Graham's Law => Solubility(S) ∝ Applied Pressure(P) => S =k·P
Given P = 0.209Atm & k = 1.28 x 10⁻³mol/L·Atm
=> S = k·P = (1.28 x 10⁻³ mole/L·Atm)0.209Atm = 2.68 x 10⁻³ mol O₂/L water.
∴Solubility of O₂(g) in 4L water at 0.209Atm = (2.68 x 10⁻³mole O₂(g)/L)(4L)(32 g O₂(g)/mol O₂(g)) = <u>3.45 x 10⁻² grams O₂(g) in 4L water. </u>