Answer:
16 mol/L
Explanation:
solution contains 8.0 moles of solute in 500 mL of solution.
so,
molarity = no. of moles of solute/volume of solution (in litres)
= 8 mol/0.5 L
=16mol/L
Answer:
In the case of mixtures of ethanol and water, this minimum occurs with 95.6% by mass of ethanol in the mixture. The boiling point of this mixture is 78.2°C, compared with the boiling point of pure ethanol at 78.5°C, and water at 100°C. You might think that this 0.3°C doesn't matter much, but it has huge implications for the separation of ethanol / water mixtures. The next diagram shows the boiling point / composition curve for ethanol / water mixtures. I've also included on the same diagram a vapor composition curve in exactly the same way as we looked at on the previous pages about phase diagrams for ideal mixtures.
Reaction of option c produces precipitate.
Rhodium on reacting with potassium phosphate produces rhodium phosphate which remain in solution due to low lattice energy for rhodium phosphate.
Niobium on reacting with lithium carbonate produces niobium carbonate and it will remain in aqueous form.
Cobalt on reacting with zinc nitrate produces cobalt nitrate. This, Co(NO3 )2 is insoluble precipitate and settles at bottom whereas zinc ion will remain in solution as follows:
Potassium ion on reacting with sodium sulfide produces potassium sulfide which remain in solution
Answer:
Mg(s) + 2NaF (aq) —> MgF₂ (aq) + 2Na(s)
Explanation:
Magnesium => Mg
Sodium fluoride => NaF
Magnesium fluoride => MgF₂
Sodium metal => Na
The equation can be written as:
Mg + NaF —> MgF₂ + Na
Thus, the above equation can be balance as illustrated below:
Mg + NaF —> MgF₂ + Na
There are 2 atoms of F on the right side and 1 atom on the left side. It can be balance by writing 2 before NaF as shown below:
Mg + 2NaF —> MgF₂ + Na
There are 2 atoms of Na on the left side and 1 atom on the right side. It can be balance by writing 2 before Na as shown below:
Mg + 2NaF —> MgF₂ + 2Na
Mg(s) + 2NaF (aq) —> MgF₂ (aq) + 2Na(s)
Now, the equation is balanced.