Answer:
1- Yes, we can calculate the solubility of mineral compound X.
2- 0.012 g/mL.
Explanation:
<em>1- Using only the information above, can you calculate the solubility of X in water at 15.0 °C? </em>
The information available is:
The volume of water sample = 25.0 mL.
Weight of the mineral compound X after evaporation, drying, and washing = 0.30 g.
∴ Yes, we can calculate the solubility of mineral compound X.
<u><em>2- If you said yes, calculate it.</em></u>
∵ 25.0 mL of water sample contains → 0.30 g of the mineral compound X.
∴ 1.0 mL of water sample contains → ??? g of the mineral compound X.
1.0 ml of water sample will contain (0.3 g/25.0 mL) 0.012 g.
<em>∴ The solubility of the mineral compound X in the water sample is</em> <u><em>0.012 g/mL.</em></u>
<u><em></em></u>
Atomic mass unit is the answer
Explanation:
This is correct!
Ions that exist in both the reactant and product side of the equation are referred to as spectator ions. Overall, they do not partake in the reaction. If they are present on both sides of the equation, you can cancel them out.
An example is;
Na+(aq) + Cl−(aq) + Ag+(aq) + NO3−(aq) → Na+(aq) + NO3−(aq) + AgCl(s)
The ions; Na+, NO3−(aq) would be cancelled out to give;
Cl−(aq) + Ag+(aq) → AgCl(s)
Water is soluting the suger WATER