We use the osmotic pressure to determine the concentration of the solute in the solution. Then, we multiply the volume of the solution to determine the number of moles of solute particles. We need to establish to equations since we have two unknowns, the mass of of each solute. We do as follows:
osmotic pressure = CRT
<span>C = 7.75 / 0.08205 (296.15) = 0.3189 mol / L</span>
<span>moles of particles = C*V = 0.3189*0.250 =0.0797 mol </span>
<span>0.0797 = moles of sucrose + 2*moles of salt </span>
<span>x + 2y = 0.0797 </span>
<span>and </span>
<span>x(MMsucrose) + y(MMNaCl) = 10.2</span>
<span>342x + 58.5y = 10.2
</span>
<span>solve for x and y
</span>
<span>x = 0.0252 mol sucrose</span>
<span>y = 0.0273 mol NaCl
</span>
<span>mass Sucrose = 0.0252(342) = 8.6184 g </span>
<span>mass NaCl = 0.0273(58.5) = 1.5971 g </span>
<span>% NaCl = (1.5971 / 10.2)*100 = 15.66%</span>
Answer:
15.Potassium oxide
16.Calcium chloride
17.Aluminium sulphide
18.CaS
Explanation:
15.K is the chemical symbol of Potassium and generally the name of the non-metal at the end of a formula has the suffix '-ide' and since O is oxygen, the name becomes Potassium oxide.
16. The same applies here. Ca is Calcium and Cl is Chlorine but since its the non-metal at the end, it ends in -ide. So Calcium chloride.
17.The same applies here too. Al is Aluminium and S is Sulphur so Aluminium sulphide.
18. Calcium's symbol is Ca and that of Sulphur is S and that gives the formula CaS.
Answer:
Explanation:
Hello.
In this case, for the reaction between aqueous solutions of ammonium chloride and iron (III) hydroxide, we have the following complete molecular reaction:
And the full ionic equation, taking into account that the iron (III) hydroxide cannot be dissolved as it is insoluble in water:
Finally, the net ionic equation, considering that spectator ions are NH₄⁺, Cl⁻ as they are both the left and right side, therefore, the net ionic equation is:
Best regards.
Answer: electronic configuration
Explanation:
Answer: similar, energy
Explanation: (edmentum answer) Several of the elements in period 5 have electron configurations with only one electron in the 5s sublevel. That suggests that the next sublevel that electrons typically fill, 4d, is not much higher in energy.
