Answer:
In 4.5 grams of tetraphosphorus decoxide we have 3.85 * 10^22 phosphorus atoms
Explanation:
Step 1: Data given
tetraphosphorus decoxide = P4O10
Molar mass of P4O10 = 283.89 g/mol
Mass of P4O10 = 4.5 grams
Number of Avogadro = 6.022 * 10^23 / mol
Step 2: Calculate moles of P4O10
Moles P4O10 = mass P4O10 / molar mass P4O10
Moles P4O10 = 4.5 grams / 283.89 g/mol
Moles = 0.016 moles
Step 3: Calculate moles of P
For 1 mol P4O10 we have 4 moles of phosphorus
For 0.016 moles P4O10 we have 4*0.016 = 0.064 moles P
Step 4: Calculate number of P atoms
Number of P atoms = moles P * number of Avogadro
Number of P atoms = 0.064 moles * 6.022*10^23
Number of P atoms = 3.85 * 10^22 atoms
In 4.5 grams of tetraphosphorus decoxide we have 3.85 * 10^22 phosphorus atoms
Answer:
Ksp = 8.8x10⁻⁵
Explanation:
<em>Full question is:</em>
<em>After mixing an excess PbCl2 with a fixed amount of water, it is found that the equilibrium concentration of Pb2+ is 2.8 × 10–2 M. What is Ksp for PbCl2?</em>
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When an excess of PbCl₂ is added to water, Pb²⁺ and Cl⁻ ions are produced following Ksp equilibrium:
PbCl₂(s) ⇄ Pb²⁺ + 2Cl⁻
Ksp = [Pb²⁺] [Cl⁻]²
If an excess of PbCl₂ was added, an amount of Pb²⁺ is produced (X) and twice Pb²⁺ is produced as Cl⁻ (2X):
Ksp = [X] [2X]²
Ksp = 4X³
As X is the amount of Pb²⁺ = 2.8x10⁻²M:
Ksp = 4(2.8x10⁻²)³
<h3>Ksp = 8.8x10⁻⁵</h3>
P1V1 = P2V2
P1 = 720 mmHg
V1 = 450. mL
P2 = 760 mmHg (this is the pressure at STP)
Use these to solve for V2:
(720)(450) = 760V2
V2 = 426 mL
Answer:
The second ring in an atom can only hold up to 8 electrons.
Let us check each statement one by one
a) Sb has a lower ionization energy but a higher electronegativity than I. : As per values given : Definitely Sb has lower ionization energy however the electronegativity of Sb is lower than that of iodine
b) Sb has a higher ionization energy but a lower electronegativity than I. FAlse:
Sb has lower ionization energy than I
c) Sb has a lower ionization energy and a lower electronegativity than I. True
d) Sb has a higher ionization energy and a higher electronegativity than I. False
