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boyakko [2]
4 years ago
10

In a campfire, carbon from wood reacts with oxygen in a combustion reaction. If you’ve ever started a campfire, you know you mus

t begin by lighting small pieces of wood (called kindling) instead of logs. Likewise, as this video shows, iron filings burn in the presence of air to make iron oxide. Why do iron filings burn while a heavy iron pipe does not? Why do twigs ignite better than logs? Give your answer in terms of atoms.
Chemistry
2 answers:
hichkok12 [17]4 years ago
7 0
Increased surface area means more<span> places where </span>wood can<span> turn into flame - where heat, fuel, and oxygen </span>are<span> all in the same place at the same time. </span>The mass of a small particle is such that the entire particle can be heated<span> to ignition temperature. An </span>iron<span> rod will </span>not burn<span> as it has relatively very less surface area per unit mass </span>but iron filings<span> with more surface area per unit mass can </span>burn<span> in air and get converted to </span>iron oxide.<span>Kindling </span>or<span> smaller pieces </span>or wood <span>have a lot of surface area and therefore have </span>more<span> contact with oxygen, making them easier to burn.</span>
Alenkasestr [34]4 years ago
5 0
One thing I can tell is the fact that more surface area means more wood to burn into flames. When it starts to burn Heat, fuel, and oxygen occur in the same place at the same time to put it into flames.




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At a certain temperature, 0.800 mol SO 3 is placed in a 1.50 L container. 2 SO 3 ( g ) − ⇀ ↽ − 2 SO 2 ( g ) + O 2 ( g ) At equil
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Answer : The value of equilibrium constant (K) is, 0.004

Explanation :

First we have to calculate the concentration of SO_3\text{ and }O_2

\text{Concentration of }SO_3=\frac{\text{Moles of }SO_3}{\text{Volume of solution}}=\frac{0.800mol}{1.50L}=1.2M

and,

\text{Concentration of }O_2=\frac{\text{Moles of }O_2}{\text{Volume of solution}}=\frac{0.150mol}{1.50L}=0.1M

Now we have to calculate the value of equilibrium constant (K).

The given chemical reaction is:

                       2SO_3(g)\rightarrow 2SO_2(g)+O_2(g)

Initial conc.       1.2             0               0

At eqm.          (1.2-2x)        2x               x

As we are given:

Concentration of O_2 at equilibrium = x = 0.1 M

The expression for equilibrium constant is:

K_c=\frac{[SO_2]^2[O_2]}{[SO_3]^2}

Now put all the given values in this expression, we get:

K_c=\frac{(2x)^2\times (x)}{(1.2-2x)^2}

K_c=\frac{(2\times 0.1)^2\times (0.1)}{(1.2-2\times 0.1)^2}

K_c=0.004

Thus, the value of equilibrium constant (K) is, 0.004

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