N2H4
<span>Each nitrogen weighs 14.01 and each H weighs 1.01. !4.01+14.01+1.01+1.01 = 32.06 (roughly) </span>
Answer:
The correct answer is A) Pb(C₂H₃O₂)₂ + Li₂SO₄
Explanation:
- All salts of Na, K and ammonium are soluble.
- All nitrates are soluble.
- All chlorides are soluble, with the exception of AgCl, Hg₂Cl₂, PbCl₂ and CuCl.
- All sulfates are soluble, with the exception of CaSO₄, SrSO₄, BaSO₄, PbSO₄, HgSO₄, Hg₂SO₄ and Ag₂SO₄.
- All hydroxides are poorly soluble, with the exception of alkaline hydroxides, Sr(OH)₂, Ba(OH)₂ and NH4(OH)
- All carbonates are poorly soluble, with the exception of alkaline carbonates and ammonium carbonate.
- All sulfides are poorly soluble, with the exception of alkaline sulfides, alkaline earth sulfides, and ammonium sulfide.
The reaction A is:
Pb(C₂H₃O₂)₂ + Li₂SO₄ ⇒PbSO₄ + 2 LiC₂H₃O₂
Lead sulfate is a slightly soluble solid, therefore it precipitates.
Here we have to choose the molecule which can generate ion-dipole interaction with ammonia.
The NH₃ will have ion dipole interaction only with b) NaOH.
a) The OCl₂ molecule is highly unstable and remains in a linear state as shown in the figure.
b) The NaOH is purely an ionic compound. There is one lone pair of electron on the nitrogen (N) atom of ammonia and also the N-H bond has dipole moment. Now among the given molecules only NaOH is ionic in nature which remains as Na⁺ and OH⁻. Thus we can expect an ion-dipole interaction between these molecules, it is shown in the figure.
c) SiO₂ is purely an covalent compound the structure is shown in the figure.
d) Methyl iodide (CH₃I) is also a covalent compound.
e) The butanol (C₄H₉OH) is a covalent compound.
So we may expect only ion-dipole interaction with b) NaOH.
Answer:
C₆H₆O
Explanation:
The given values are the percentage in mass of each element. Empirical formula is the simplest ratio of atoms presents in a molecule. Thus, the first we need to do is convert this percentage to mass using molar mass of each element:
<em>Moles C:</em>
76.57C * (1mol / 12g) = 6.38 moles C
<em>Moles H:</em>
6.43 * (1mol / 1g) = 6.43 moles H
<em>Moles O:</em>
17.0% * (1mol / 16g) = 1.06 moles O
Dividing each number of moles in the moles of O (The minimum number of moles. Used to obtain "The simplest ratio..."):
C = 6.38 moles C / 1.06 moles O = 6
H = 6.43 moles H / 1.06 moles O = 6
O = 1.06 moles O / 1.06 moles O = 1
That means empirical formula of phenol is:
<h3>C₆H₆O</h3>
