Answer:
36365.4 Joules
Explanation:
The quantity of Heat Energy (Q) released on cooling a heated substance depends on its Mass (M), specific heat capacity (C), and change in temperature (Φ)
Thus, Q = MCΦ
Since, M = 45.4 g
C = 3.56 J/g°C,
Φ = 250°C - 25°C = 225°C
Q = 45.4g x 3.56J/g°C x 225°C
Q= 36365.4 Joules
Thus, 36365.4 Joules of heat energy is released when the lithium is cooled.
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Answer:
Different Number of neutrons.
The mass of reacted magnesium chloride is 23.75g, percent by mass of solution magnesium chloride is 90.9%.
<h3>What is the relation between mass & moles?</h3>
Relation between the mass and moles of any substance will be represented as:
n = W/M, where
- W = given mass
- M = molar mass
Moles of MgCl₂ = 200g / 95g/mol = 2.1mol
Moles of NaOH = 20g / 0.5mol
Given chemical reaction is:
MgCl₂ + 2NaOH → 2NaCl + Mg(OH)₂
From the stoichiometry of the reaction it is clear that
- 1 mole of MgCl₂ = reacts with 2 moles of NaOH
- 0.5 mole of NaOH = reacts with (1/2)(0.5)=0.25 moles of MgCl₂
Mass of reacted MgCl₂ = (0.25mol)(95g/mol) = 23.75g
Percent by mass of MgCl₂ in the given solution mixture will be calculated as:
- % by mass = (Mass of MgCl₂ / Total mass of mixture) × 100%
- Percent by mass of MgCl₂ = (200/200+20)×100% = 90.0&
In the above reaction we obtain NaCl as a solid, and MgCl₂ is the limiting reagent in it, from which 2 moles of NaCl is produced means
- 0.25 moles of MgCl₂ = produces 0.5 moles of NaCl
Mass of NaCl = (0.5mol)(58.5g/mol) = 29.25g
Hence, mass of reacted MgCl₂ is 23.75g, percent by mass of solution magnesium chloride is 90.9% and mass of the obtained solid is 29.25g.
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Answer:
2Al(s) +3Ni²⁺(aq) ⟶ 2Al³⁺(aq) + 3Ni(s)
Explanation:
The unbalanced equation is
Al(s) + Ni²⁺(aq) ⟶ Ni(s) + Al³⁺(aq)
(i) Half-reactions
Al(s) ⟶ Al³⁺(aq) + 3e⁻
Ni²⁺(aq) + 2e⁻ ⟶ Ni(s)
(ii) Balance charges
2 × [Al(s) ⟶ Al³⁺(aq) + 3e⁻]
3 × [Ni²⁺(aq) + 2e⁻ ⟶ Ni(s)]
gives
2Al(s) ⟶ 2Al³⁺(aq) + 6e⁻
3Ni²⁺(aq) + 6e⁻ ⟶ 3Ni(s)
(iii) Add equations
2Al(s) ⟶ 2Al³⁺(aq) + 6e⁻
<u>3Ni²⁺(aq) + 6e⁻ ⟶ 3Ni(s) </u>
2Al(s) +3Ni²⁺(aq) + <em>6e</em>⁻ ⟶ 2Al³⁺(aq) + 3Ni(s) + <em>6e⁻
</em>
Simplify (cancel electrons)
2Al(s) +3Ni²⁺(aq) ⟶ 2Al³⁺(aq) + 3Ni(s)
