Answer:
4.285 L of water must be added.
Explanation:
Hello there!
In this case, for this dilution-like problems, we need to figure out the final volume of the resulting solution so that we would be able to obtain the correct volume of diluent (water) to be added. In such a way, we can obtain the final volume, V2, as shown below:
Thus, by plugging in the initial molarity, initial volume and final molarity (0.587 M) we obtain:
It means we need to add:
Of diluent water.
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Answer:
ΔH°rxn = - 433.1 KJ/mol
Explanation:
- CH4(g) + 4Cl2(g) → CCl4(g) + 4HCl(g)
⇒ ΔH°rxn = 4ΔH°HCl(g) + ΔH°CCl4(g) - 4ΔH°Cl2(g) - ΔH°CH4(g)
∴ ΔH°Cl2(g) = 0 KJ/mol.....pure element in its reference state
∴ ΔH°CCl4(g) = - 138.7 KJ/mol
∴ ΔH°HCl(g) = - 92.3 KJ/mol
∴ ΔH°CH4(g) = - 74.8 KJ/mol
⇒ ΔH°rxn = 4(- 92.3 KJ/mol) + (- 138.7 KJ/mol) - 4(0 KJ/mol) - (- 74.8 KJ/mol)
⇒ ΔH°rxn = - 369.2 KJ/mol - 138.7 KJ/mol - 0 KJ/mol + 74.8 KJ/mol
⇒ ΔH°rxn = - 433.1 KJ/mol
There are a lot of ways to increase the solubility of the solute. <span>Increasing the temperature, mixing time and surface area of a solvent increases the solubility of the solute</span>
Answer:
Resolution is the ability to distinguish two objects from each other. Light microscopy has limits to both its resolution and its magnification.
