The pH = 2.41
<h3>Further explanation</h3>
Given
5.0% by mass solution of acetic acid
the density of white vinegar is 1.007 g/cm3
Required
pH
Solution
Molarity of solution :

Ka for acetic acid = 1.8 x 10⁻⁵
[H⁺] for weak acid :
![\tt [H^+]=\sqrt{Ka.M}](https://tex.z-dn.net/?f=%5Ctt%20%5BH%5E%2B%5D%3D%5Csqrt%7BKa.M%7D)
Input the value :
![\tt [H^+]=\sqrt{1.8\times 10^{-5}\times 0.839}\\\\(H^+]=0.00388=3.88\times 10^{-3}\\\\pH=3-log~3.88=2.41](https://tex.z-dn.net/?f=%5Ctt%20%5BH%5E%2B%5D%3D%5Csqrt%7B1.8%5Ctimes%2010%5E%7B-5%7D%5Ctimes%200.839%7D%5C%5C%5C%5C%28H%5E%2B%5D%3D0.00388%3D3.88%5Ctimes%2010%5E%7B-3%7D%5C%5C%5C%5CpH%3D3-log~3.88%3D2.41)
Answer:
This description needs a unit describing the system of measurement.
Answer:
1) When 6.97 grams of sodium(s) react with excess water(l), 56.0 kJ of energy are evolved.
2) When 10.4 grams of carbon monoxide(g) react with excess water(l), 1.04 kJ of energy are absorbed.
Explanation:
1) The following thermochemical equation is for the reaction of sodium(s) with water(l) to form sodium hydroxide(aq) and hydrogen(g).
2 Na(s) + 2H₂O(l) ⇒ 2NaOH(aq) + H₂(g) ΔH = -369 kJ
The enthalpy of the reaction is negative, which means that 369 kJ of energy are evolved per 2 moles of sodium. The energy evolved for 6.97 g of Na (molar mass 22.98 g/mol) is:

2) The following thermochemical equation is for the reaction of carbon monoxide(g) with water(l) to form carbon dioxide(g) and hydrogen(g).
CO(g) + H₂O(l) ⇒ CO₂(g) + H₂(g) ΔH = 2.80 kJ
The enthalpy of the reaction is positive, which means that 2.80 kJ of energy are absorbed per mole of carbon monoxide. The energy evolved for 10.4 g of CO (molar mass 28.01 g/mol) is:

Those reactions in which Alkyl Halide reacts with the solvent without the involvement of any acid or base is called as
Solvolysis. In given problem <em>tert</em>-Butyl Bromide is a tertiary Alkyl Halide and we know well that tertiary alkyl halides undergo
SN¹ and
E¹ elimination reaction due to the formation of
stable tertiary carbocation. In given example after the formation of carbocation when Isopropyl act as
nucleophile it will produce
ether and when it acts as a
base it will produce
unsaturated compound. The reaction along with both products is shown below,