First convert to moles:
0.5113 g / 17 g/mol = 0.0301 mol
Now create a ratio based on the reaction provided to solve for the unknown:
4 NH3 / -905.4 kJ = 0.0301 mol NH3 / x kJ
x = -6.808 kJ
Mass of water produced : 0.146 g
<h3>Further explanation</h3>
Given
33.2 mL of 0.245 M lithium hydroxide
Required
mass of water
Solution
Reaction
HNO₃ (aq) + LiOH (aq) → H₂O (l) + LiNO₃ (aq)
mol LiOH :
= M x V
= 0.245 x 33.2 ml
= 8.134 mmol
From the equation, the mol ratio of HNO₃ : H₂O = 1 : 1, so mol H₂O = 8.134 mmol
mass H₂O :
= mol x MW
= 8.134 x 10⁻³ mol x 18 g/mol
= 0.146 g
Answer:
Change in internal energy (ΔU) = -9 KJ
Explanation:
Given:
q = –8 kJ [Heat removed]
w = –1 kJ [Work done]
Find:
Change in internal energy (ΔU)
Computation:
Change in internal energy (ΔU) = q + w
Change in internal energy (ΔU) = -8 KJ + (-1 KJ)
Change in internal energy (ΔU) = -8 KJ - 1 KJ
Change in internal energy (ΔU) = -9 KJ
Molality is the number of moles of solutes in 1 kg of solvent.
the molality of solution to be prepared is 2.0 molal.
therefore 2 moles in 1 kg water.
the mass of Li₂S required is - 46 g/mol x 2.0 mol = 92 g
the mass in 1 kg of solvent is - 92 g
Therefore mass of Li₂S required in 1600.0 g is - 92 g/kg x 1.6 kg = 147.2 g
