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Sphinxa [80]
3 years ago
12

the molar enthalpy of formation of carbondioxide is -393kjmol. calculate the heat released by the burning of 0.327g of carbon to

form carbondioxide.
Chemistry
1 answer:
Arte-miy333 [17]3 years ago
7 0

This is equivalent to having a standard enthalpy change of reaction equal to  10.611 kJ

<u>Explanation</u>:

The standard enthalpy change of reaction,  Δ H ∘ , is given to you in kilojoules per mole, which means that it corresponds to the formation of one mole of carbon dioxide.

                                    C (s]  +  O 2(g] → CO 2(g]

Remember, a negative enthalpy change of reaction tells you that heat is being given off, i.e. the reaction is exothermic.

First to convert  grams of carbon into moles,

use carbon's molar mass(12.011 g).

                    Moles of C = mass in gram / molar mass

                                        = 0.327 g  / 12.011 g

                     Moles of C = 0.027 moles

Now, in order to determine how much heat is released by burning of 0.027 moles of carbon to form carbon-dioxide.

                                        =  0.027 moles C  \times 393 kJ

             Heat released  = 10.611  kJ.

So, when  0.027  moles of carbon react with enough oxygen gas, the reaction will give off  10.611 kJ  of heat.

This is equivalent to having a standard enthalpy change of reaction equal to  10.611 kJ

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<span>For two situations, this law can be added as;
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3 0
3 years ago
guys please help me WILL MARK BARINLIEST If you have 5.7 x 10^24 molecules of O2 at STP, how many liters of O2 do you have? must
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Answer:

212.8 dm^3 or L

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Answer:

0.011 mol/L

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I = initial

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x amount of N₂ reacts.  Since the stoichiometry is 1:1, x amount of O₂ also reacts.  This produces 2x of NO.

After the reaction, there is 0.04-x of N₂, 0.04-x of O₂, and 2x of NO.

Here it is in table form:

\left[\begin{array}{cccc}&N2&O2&NO\\I&0.04&0.04&0\\C&-x&-x&+2x\\E&0.04-x&0.04-x&2x\end{array}\right]

Now we can use the equilibrium constant:

Kc = [NO]² / ( [N₂] [O₂] )

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