The Hbr has highest boiling point followed by Kr , 
and 
.
The Hbr has highest boiling point due to presence of intermolecular H-bonding that is hydrogen bonding.
The electronegativity of bromine is greater than hydrogen atom. The electronegativity difference hydrogen atom and bromine atom is high. So, due to this a positive dipole is generated at hydrogen atom and partial negative charge, these partial charge is called dipole. In a HBr molecule one H-atom attracted by Br-atom of another atom by force of attraction and this partial force of attraction is called H-bonding.
Other all given molecules are arrange according to its molecular mass because boiling points is directly proportional to the molecular mass of molecule.
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Answer:
A. 28 years
Explanation:
Applying,
R = R'(2ᵃ/ⁿ).............. Equation 1
Where R = Original sample, R' = Sample left after decay, a = Total time taken to decay, n = half life.
From the question,
Given: R = 12 g, R' = 6 g, a = 28 years.
Substitute into equation 1 and solve for n
12 = 6(2²⁸/ⁿ)
12/6 = 2²⁸/ⁿ
2²⁸/ⁿ = 2
Equation the base,
28/n = 1
n = 28 years.
Hence the half-life is 28 years
As mentioned above, phosphoric acid has 3 pKa values, and after 3 ionization it gives 3 types of ions at different pKa values:
H₃PO₄(aq)
+ H₂O(l) ⇌ H₃O⁺(aq) + H₂PO₄⁻ (aq) pKₐ₁
<span>
</span>H₂PO₄⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + HPO₄²⁻ (aq) pKₐ₂
HPO₄²⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + PO₄³⁻ (aq) pKₐ₃
At the highest pKa value (12.4) of phosphoric acid, the last OH group will lose its hydrogen. On the picture I attached, it is shown required protonated form of phosphoric acid before reaction whose pKa value is 12.4.
<span>The composition of a fertilizer is usually express in NPK number. NPK number is in terms of Percent by mass of the said element which are Nitrogen, Phosphorus and Potassium. A 15-35-15 fertilizer has 15%
Nitrogen, 35% Phosphorous, and 15% Potassium by mass. If you have 10 g of this
fertilizer, to get the number of moles of phosphorus, you multiply the mass by
35%, which is equal to 10*0.35 or 3.5 g phosphorus. Then you divide the
calculated mass of phosphorous by its molar mass which is 30.97 g/mol.
Therefore, you have 3.5/30.97 which is equal to 0.1130 mol Phosphorus. This is the amount of Phosphorus in moles in the fertilizer.</span>
