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Svet_ta [14]
3 years ago
10

Palmitic acid C16H32O2 is a dietary fat found in beef and butter. The caloric content of palmitic acid is typical of fats in gen

eral.
Part A
Write a balanced equation for the complete combustion of palmitic acid. Use \rm H_2O(l) in the balanced chemical equation because the metabolism of these compounds produces liquid water.
Express your answer as a chemical equation. Identify all of the phases in your answer.
Chemistry
2 answers:
Luden [163]3 years ago
7 0
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) 

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
</span>
egoroff_w [7]3 years ago
6 0

Explanation:

Combustion is a type of chemical reaction in which substance burns under the presence of oxygen or in other words reaction of hydrocarbon with oxygen to produce water and carbondioxide.

Palmitic acid when undergoes combustion it gives carbondioxide and water.

The balanced chemical  equation is given as:

C_{16}H_{32}O_2(s)+23O_2(g)\rightarrow 16CO_2(g)+16H_2O(l)

According to stoichiometry, when 1 mol of palmitic acid reacts with 23 moles of oxygen to give 16 moles of carbondioxide and 12 moles of water.

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4 years ago
The decomposition of N_2O_5(g) following 1st order kinetics. N_2O_5(g) to N_2O_4(g) + ½ O_2(g) If 2.56 mg of N_2O_5 is initially
Crank

Answer: The rate constant is 0.334s^{-1}

Explanation ;

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = ?

t = age of sample  = 4.26 min

a =  initial amount of the reactant  = 2.56 mg

a - x = amount left after decay process  = 2.50 mg

Now put all the given values in above equation to calculate the rate constant ,we get

4.26=\frac{2.303}{k}\log\frac{2.56}{2.50}

k=\frac{2.303}{4.26}\log\frac{2.56}{2.50}

k=5.57\times 10^{-3}min^{-1}=5.57\times 10^{-3}\times 60s^{-1}=0.334s^{-1}

Thus rate constant is [tex]0.334s^{-1}

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