Answer:
Option 4. There's no hydrogen bonding between HBr molecules at all.
Explanation:
<h3>SiH₄</h3>
SiH₄ molecules are tetrahedral and symmetric. Dipoles due to the polar Si-H bonds balance each other. SiH₄ molecules are nonpolar. Only instantaneous dipoles are possible between those molecules.
<h3>C₆H₆ Benzene</h3>
Similar to SiH₄, benzene is symmetric. Dipoles due to the weakly polar C-H bonds balance each other. Benzene molecules are nonpolar. Only instantaneous dipoles are possible between those molecules.
<h3>NH₃</h3>
There are two conditions for hydrogen bonding to take place:
- H atoms are directly bonded to a highly electronegative element: Nitrogen, Oxygen, or Fluorine.
- There is at least one lone pair of electrons nearby.
Consider the Lewis structure of NH₃. There are three H atoms in each NH₃ molecule. Each of the three H atoms is bonded directly to the N atom with a highly polar N-H bond. Also, there is a lone pair of electrons on the N atom. Hydrogen bonding will take place between NH₃ molecules.
NH₃ is a relatively small molecule. As a result, hydrogen bonding will be the dominant type of intermolecular force between NH₃ molecules.
<h3>HBr</h3>
There are three lone pairs on the Br atom in each HBr molecule. However, no H atom is connected to any one of the three highly electronegative elements: N, O, or F. The Br atom isn't electronegative enough for the H atom to form hydrogen bonding. HBr molecules are polar. As a result, the dominant type of intermolecular forces between HBr molecules will be dipole-dipole interactions (A.k.a. permanent dipole.)
<h3>CaO</h3>
Calcium is a group 2 metal. Oxygen is one of the three most electronegative nonmetal. (Again, the most electronegative elements are: Nitrogen, Oxygen, and Fluorine.) As a main group metal, Ca atoms tend to lose electrons and form positive ions. Oxygen will gain those electrons to form a negative ion. As a result, CaO will be an ionic compound full of Ca²⁺ and O²⁻ ions. Forces between ions with opposite charges are called ionic bonds.
Answer:
a) pH = 4.68 (more effective)
b) pH =4.44.
Explanation:
The pH of buffer solution is obtained by Henderson Hassalbalch's equation.
The equation is:
![pH =pKa +log\frac{[salt]}{[acid]}](https://tex.z-dn.net/?f=pH%20%3DpKa%20%2Blog%5Cfrac%7B%5Bsalt%5D%7D%7B%5Bacid%5D%7D)
a) pKa of acetic acid = 4.74
[salt] = [CH₃COONa] = 1.4 M
[acid] = [CH₃COOH] = 1.6 M
This is more effective as there is very less difference in the concentration of salt and acid.
b) pKa of acetic acid = 4.74
[salt] = [CH₃COONa] = 0.1 M
[acid] = [CH₃COOH] = 0.2 M
Methanol is prepared by reacting Carbon monoxide and Hydrogen gas,
CO + 2 H₂ → CH₃OH
Calculating Moles of CO:
According to equation,
32 g (1 mole) of CH₃OH is produced by = 1 Mole of CO
So,
3.60 × 10² g of CH₃OH is produced by = X Moles of CO
Solving for X,
X = (3.60 × 10² g × 1 Mole) ÷ 32 g
X = 11.25 Moles of CO
Calculating Moles of H₂:
According to equation,
32 g (1 mole) of CH₃OH is produced by = 2 Mole of H₂
So,
3.60 × 10² g of CH₃OH is produced by = X Moles of H₂
Solving for X,
X = (3.60 × 10² g × 2 Mole) ÷ 32 g
X = 22.5 Moles of H₂
Result:
3.60 × 10² g of CH₃OH is produced by reacting 11.25 Moles of CO and 22.5 Moles of H₂.
Hi, here is a basic summary of what we did in a lab; there were 3 reactions: The procedure: Reaction 1: Solid sodium hydroxide dissolves in water to form an aqueous solution of ions. NaOH(s)-> Na+(aq) + OH-(aq) ΔH1=-34.121kJ Reaction 2: Solid sodium hydroxide reacts with an aqueous solution of HCl to form water and an aqueous solution of sodium chloride. NaOH(s) + H+(aq) + Cl-(aq) -> H2O + Na+(aq) + Cl-(aq) ΔH2=-83.602kJ Reaction 3: An aqueous solution of sodium hydroxide reacts with an aqueous solution of HCl to form water an an aqueous solution of sodium chloride. H+(aq) + OH-(aq) + Na+(aq) + Cl-(aq) -> H2O + Na+(aq) + Cl-(aq) ΔH3= -50.2kJ The ΔH values were calculated by dividing the heat gained by the number of moles (each reaction had 0.05moles of NaOH) The problem: Net ionic equations for reaction 2 & 3: 2: NaOH(s) + H+(aq) -> H2O + Na+(aq) 3: H+(aq) + OH-(aq) -> H2O i) In reaction 1, ΔH1 represents the heat evolved as solid NaOH dissolves. Look at the net ionic equations for reactions 2 and 3 and make similar statements as to what ΔH2 and ΔH3 represent. ii) Compare ΔH2 with (ΔH1 + ΔH3). Explain in sentences the similarity between these two values by using your answer to #5 above. Attempt at answering: i) Firstly, ΔH2 represents the heat evolved as the hydrogen ion displaces the sodium ion, creating a single displacement reaction. ΔH3 represents the heat evolved as the hydrogen and hydroxide ion form water via a neutralization reaction. ii) ΔH2 is equal to (or supposed to be, this is a source of error while calculating) (ΔH1 + ΔH3). The similarity between these two values is that .. (this is where I get confused!)
Source https://www.physicsforums.com/threads/calorimetry-help-chemistry.399653/
