Answer:
See explanation
Explanation:
The boiling point of a substance is affected by the nature of bonding in the molecule as well as the nature of intermolecular forces between molecules of the substance.
2-methylpropane has only pure covalent and nonpolar C-C and C-H bonds. As a result of this, the molecule is nonpolar and the only intermolecular forces present are weak dispersion forces. Therefore, 2-methylpropane has a very low boiling point.
As for 2-iodo-2-methylpropane, there is a polar C-I bond. This now implies that the intermolecular forces present are both dispersion forces and dipole interaction. As a result of the presence of stronger dipole interaction between 2-iodo-2-methylpropane molecules, the compound has a higher boiling point than 2-methylpropane.
Answer: B) Particles can be filtered from a suspension.
Explanation: Colloids are solutions with particle size intermediate between true solutions and suspensions. They exhibit tyndall effect that is scattering of light.
Suspensions have large sized particles which settle when left undisturbed for sometime and thus can be filtered off easily.
The particle size in colloids is less and hence they do not settle under the effect of gravity.
A solution can be homogeneous in which the composition is uniform or heterogeneous in which is the composition is not uniform.
In this item, we are simply to find the ions that may bond and are able to form a formula unit. We are also instructed to give out their name. There are numerous possible combinations of ions to form a compound. Some answers are given in the list below.
1. Na⁺ , Cl⁻ , NaCl ---> sodium chloride (this is most commonly known as table salt)
2. C⁴⁺ , O²⁻ , CO₂ ---> carbon dioxide
3. Al³+ , Cl⁻ , AlCl₃ ----> aluminum chloride
4. Ca²⁺ , Cl⁻ , CaCl₂ ---> calcium chloride
5. Li⁺ , Br⁻ , LiBr ---> lithium bromide
6. Mg³⁺ , O²⁻ , Mg₂O₃ ----> magnesium oxide
7. K⁺ , I⁻ , KI ---> potassium iodide
8. H⁺ , Cl⁻ , HCl --> hydrogen chloride
9. H⁺ , Br⁻ , HBr ----> hydrogen bromide
10. Na⁺ , Br⁻ , NaBr ---> sodium bromide
<u>Answer:</u> The initial pH of the HCl solution is 3
<u>Explanation:</u>
To calculate the concentration of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is HCl
are the n-factor, molarity and volume of base which is NaOH.
We are given:
Putting values in above equation, we get:
1 mole of HCl produces 1 mole of 
ions and 1 mole of 
ions
To calculate the pH of the solution, we use the equation:
![pH=-\log[H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D)
We are given:
![[H^+]=0.001M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.001M)
Putting values in above equation, we get:
Hence, the initial pH of the HCl solution is 3
