Answer:
Adiabatic. This is a process where no heat is being added or removed from the system. Or can be simply stated as: no heat transfer (or heat flow) happening in a system.
Explanation:
We have to know the molarity of solution obtained when 5.71 g of Na₂CO₃.10 H₂O is dissolved in water and made up to 250 cm³ solution.
The molarity of solution obtained when 5.71 g of sodium carbonate-10-water (Na₂CO₃.10 H₂O) is dissolved in water and made up to 250.0 cm^3 solutionis: (A) 0.08 mol dm⁻³
The molarit y of solution means the number of moles of solute present in one litre of solution. Here solute is Na₂CO₃.10 H₂O and solvent is water. Volume of solution is 250 cm³.
Molar mass of Na₂CO₃.10 H₂O is 286 grams which means mass of one mole of Na₂CO₃.10 H₂O is 286 grams.
5.71 grams of Na₂CO₃.10 H₂O is equal to
= 0.0199 moles of Na₂CO₃.10 H₂O. So, 0.0199 moles of Na₂CO₃.10 H₂O present in 250 cm³ volume of solution.
Hence, number of moles of Na₂CO₃.10 H₂O present in one litre (equal to 1000 cm³) of solution is
= 0.0796 moles. So, the molarity of the solution is 0.0796 mol/dm³ ≅ 0.08 mol/dm³
Answer:
Explanation:
The formula for sodium is Na. It does not form a molecule in some way.
1 mol Na = 6.02*10^23 atoms
3.91 mol = x Cross multiply
x = 3.91 * 6.02 * 10^23
x = 23.65 * 10^23
x = 2.365 * 10^24
Scientific notation is always expressed as a number 1 ≤ x < 10
Molarity = moles of solute/volume of solution in liters.
From this relation, we can figure out the number of moles of solute by multiplying the molarity of the solution by the volume in liters.
We have 53.1 mL, or 0.0531 L, of a 12.5 M, or 12.5 mol/L, solution. Multiplying 12.5 mol/L by 0.0531 L, we obtain 0.664 moles. So, in this volume of solution, there are 0.664 moles of solute (HCl).