Answer:
We need 12.26 grams H2SO4
Explanation:
Step 1: Data given
Volume of a H2SO4 solution = 500 mL = 0.500 L
Concentration of the H2SO4 solution = 0.250 M
Molar mass of H2SO4 = 98.08 g/mol
Step 2: Calculate moles H2SO4
Moles H2SO4 = concentration * volume
Moles H2SO4 = 0.250 M * 0.500 L
Moles H2SO4 = 0.125 moles
Step 3: Calculate mass of H2SO4
Mass of H2SO4 = moles * molar mass
Mass of H2SO4 = 0.125 moles * 98.08 g/mol
Mass of H2SO4 = 12.26 grams
We need 12.26 grams H2SO4
Answer:
This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M
Explanation:
Step 1: Data given
Molarity of Na2CrO4 = 0.010 M
Molarity of NaBr = 2.5 M
Ksp(PbCrO4) = 1.8 * 10^–14
Ksp(PbBr2) = 6.3 * 10^–6
Step 2: The balanced equation
PbCrO4 →Pb^2+ + CrO4^2-
PbBr2 → Pb^2+ + 2Br-
Step 3: Define Ksp
Ksp PbCrO4 = [Pb^2+]*[CrO4^2-]
1.8*10^-14 = [Pb^2+] * 0.010 M
[Pb^2+] = 1.8*10^-14 /0.010
[Pb^2+] = 1.8*10^-12 M
The minimum [Pb^2+] needed to precipitate PbCrO4 is 1.8*10^-12 M
Ksp PbBr2 = [Pb^2+][Br-]²
6.3 * 10^–6 = [Pb^2+] (2.5)²
[Pb^2+] = 1*10^-6 M
The minimum [Pb^2+] needed to precipitate PbBr2 is 1*10^-6 M
This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M
Correct Answer: Option C
Reason:
<span>The </span>Pauli Exclusion Principle<span> states as '<em>in an atom or molecule, no two electrons can have the same four electronic quantum numbers. Further, an orbital can contain a maximum of only two electrons, the two electrons must have opposing spins.</em>'
</span>
Thus, it can be seen that in option C, electrons in last 2 subshell have electrons with same spin, which is a violation of Pauli Exclusion Principle .
44g of CO2 can produce by the reaction of carbon with oxygen
Answer:
Br^-1
Explanation:
It is an ion that gained an extra electron making it 36 electrons but the other. one is in the neutral state with 35 electrons
