Explanation:
First you must calculate the number of moles in this solution, by rearranging the equation. No. Moles (mol) = Molarity (M) x Volume (L) = 0.5 x 2. = 1 mol.
For NaCl, the molar mass is 58.44 g/mol. Now we can use the rearranged equation. Mass (g) = No. Moles (mol) x Molar Mass (g/mol) = 1 x 58.44. = 58.44 g.
The enthalpy change : -196.2 kJ/mol
<h3>Further explanation </h3>
The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation
The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)
(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)
The value of ° H ° can be calculated from the change in enthalpy of standard formation:
∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)
Reaction
2 H₂O₂(l)-→ 2 H₂O(l) + O₂(g)
∆H ° rxn = 2. ∆Hf ° H₂O - 2. ∆Hf °H₂O₂

Answer:
Molecular Weight
Explanation:
Chromium(III) Carbonate Cr2(CO3)3 Molecular Weight -- EndMemo.
Answer is: <span>the mass of the glucose is 81,07 grams.
</span>c(C₆H₁₂O₆) = 0,3 M = 0,3 mol/L.
V(C₆H₁₂O₆) = 1,500 L.
n(C₆H₁₂O₆) = c(C₆H₁₂O₆) · V(C₆H₁₂O₆).
n(C₆H₁₂O₆) = 0,3 mol/L · 1,5 L.
n(C₆H₁₂O₆) = 0,45 mol.
m(C₆H₁₂O₆) = n(C₆H₁₂O₆) · M(C₆H₁₂O₆).
m(C₆H₁₂O₆) = 0,45 mol · 180,156 g/mol.
m(C₆H₁₂O₆) = 81,07 g.