C16H32O2(aq) --> 16CO2(g) + 16H2O(l) ... said its wrong though?
<span>This is because you haven't added any oxygen needed for the combustion, so your equation does'nt balance. Also a solution in water [aq] doesn't burn! </span>
<span>Try </span><span>C16H32O2(s) + 23O2(g) --> 16CO2(g) + 16H2O(l)
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Answer:
Glow sticks and match would be light emission, slime would be preciptate, and cookies would be gas.
Explanation:
The enthalpy of this reaction is -5315 KJ/mol.
The equation of the reaction is;
2C4H10(g) + 13O2(g) -----> 8CO2 (g) + 10H2O(g)
We know that the enthalpy of reaction can be obtained from the enthalpy of formation of the reactants and products as follows;
ΔHrxn = ΔHf(products) - ΔHf(reactants)
We have the following information from the question;
ΔHf C4H10 = - 125. 6 kJ/mol
ΔHf CO2 = - 393. 5 kJ/mol
ΔHf H2O = - 241. 82 kJ/mol
ΔHf O2 = 0 KJ/mol
Hence;
[(8 × (- 393. 5 )) + (10 × (- 241. 82))] - [2( - 125. 6))]
= -5315 KJ/mol
Learn more: brainly.com/question/13164491
The <span>ion that is formed by providing the second ionization to remove an electron is </span><span>N3-. The answer is letter B. The rest of the choices do not answer the question above.</span>