<span>Important information to solve the exercise :
Substance ΔHf (kJ/mol):
HCl(g)= −92.0 </span><span>kJ/mol
Al(OH)3(s)= −1277.0 </span><span><span>kJ/mol
</span> H2O(l)= −285.8 </span><span>kJ/mol
AlCl3(s) =−705.6 </span><span>kJ/mol
</span><span>Al(OH)3(s)+3HCl(g)→AlCl3(s)+3H2O(l)
reactants products
products- reactants:</span><span>
(−705.6) + (3 x −285.8) - ( −1277.0 ) - (3 x −92.0 ) = - 10.0 </span>kJ per mole at 25°C
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Answer ( 3 ) :
<span>conversion of matter to energy .
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hope this helps!
Answer:
Specific heat of calcium carbonate(C) = 0.82 (Approx)
Explanation:
Given:
Energy absorbs (q) = 85 J
Change in temperature (Δt) = 34.9 - 21 = 13.9°C
Mass of calcium carbonate = 7.47 g
Find:
Specific heat of calcium carbonate(C)
Computation:
Specific heat of calcium carbonate(C) = q / m(Δt)
Specific heat of calcium carbonate(C) = 85 / (7.47)(13.9)
Specific heat of calcium carbonate(C) = 85 / 103.833
Specific heat of calcium carbonate(C) = 0.8186
Specific heat of calcium carbonate(C) = 0.82 (Approx)
The answer is studying how people use time.
Hope this helps!
N(CH₃OH)=3,62·10²⁴/6·10²³ 1/mol = 6,033 mol
m(CH₃OH) = 6,033 mol · 32 g/mol (molar mass) = 193,06 g.
