Sorry it’s messy, hope this helps :]
<u>Answer:</u> The mass of methylene blue that must be weighed is 
<u>Explanation:</u>
To calculate the mass of solute, we use the equation used to calculate the molarity of solution:
We are given:
Molarity of solution = 
Molar mass of methylene blue = 319.85 g/mol
Volume of solution = 100.0 mL
Putting values in above equation, we get:
Hence, the mass of methylene blue that must be weighed is
Answer:
A. Interactions between the ions of sodium chloride (solute-solute interactions).
B. Interactions involving dipole-dipole attractions (solvent-solvent interactions).
C. Interactions formed during hydration (solute-solvent interactions).
D. Interactions involving ion-ion attractions (solute-solute interactions).
E. Interactions associated with an exothermic process during the dissolution of sodium chloride (solute-solvent interactions).
F. Interactions between the water molecules (solvent-solvent interactions).
G. Interactions formed between the sodium ions and the oxygen atoms of water molecules (solute-solvent interactions).
Explanation:
The solution process takes place in three distinct steps:
- Step 1 is the <u>separation of solvent molecules.
</u>
- Step 2 entails the <u>separation of solute molecules.</u>
These steps require energy input to break attractive intermolecular forces; therefore, <u>they are endothermic</u>.
- Step 3 refers to the <u>mixing of solvent and solute molecules.</u> This process can be <u>exothermic or endothermic</u>.
If the solute-solvent attraction is stronger than the solvent-solvent attraction and solute-solute attraction, the solution process is favorable, or exothermic (ΔHsoln < 0). If the solute-solvent interaction is weaker than the solvent-solvent and solute-solute interactions, then the solution process is endothermic (ΔHsoln > 0).
In the dissolution of sodium chloride, this process is exothermic.
Your first step should be to analyse the compound. For example, if the compound is carbon, you know it always has a valence of four, so, if it has a formula C2H4 (ethylene) it obviously has a double bond. There are difficulties here because benzene C6H6 can be considered to have 6 1.5 C-C bonds, being aromatic.
A second step is to look at its structure. Double bonds are traditionally shorter than single bonds; triple bonds shorter still. Covalent bonds do have typical lengths, nevertheless you can still have problems.
<span>A third step is to consider reactivity. For example, if you have a C=C double bond, you can add, say, bromine to it Thus C2H4 gives C2H4Br2, and by adding two bromine atoms you know you have one double bond. Again, benzene becomes an awkward molecule, but because of this, you know benzene does not have double bonds in the traditional sense</span>
