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weeeeeb [17]
3 years ago
13

Burning a piece of wood in fire can be best described as a chemical change because the atoms in wood change their state. physica

l change because the atoms in wood are turned into a gaseous product. chemical change because the atoms in wood and oxygen are rearranged. physical change because the atoms in wood are turned into black ash.
Chemistry
2 answers:
Phoenix [80]3 years ago
6 0
The answeris c, chemicalchange because the atoms in woodand oxygen are rearranged.
Misha Larkins [42]3 years ago
6 0

Answer: chemical change because the atoms in wood and oxygen are rearranged.

Explanation: The chemical composition of wood varies from species to species, but is approximately 50% carbon, 42% oxygen, 6% hydrogen, 1% nitrogen, and 1% other elements.

When wood is being combusted in air(oxygen being the gas that rekindles the glowing splint), the chemicals are rearranged. Since about 50% of wood is carbon, C, the main chemical reaction going on here is C + O2 --> CO2. Of course, there are other chemical reactions too. But the formation of CO2 is the most common.

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Likurg_2 [28]

The final temperature of the mixture : 21.1° C  

<h3>Further explanation  </h3>

The law of conservation of energy can be applied to heat changes, i.e. the heat received / absorbed is the same as the heat released  

Q in(gained) = Q out(lost)  

Heat can be calculated using the formula:  

Q = mc∆T  

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c = specific heat, joules / g ° C  

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Q ethanol=Q water

mass ethanol=

\tt mass=\rho\times V\\\\mass=0.789\times 45=35.505~g

mass water =

\tt mass=1~g/ml\times 45~ml=45~g

then the heat transfer :

\tt 35.505\times 2.42~J/g^oC\times (t-6)=45\times 4.18~J/g^oC\times (28-t)\\\\85.922t-515.533=5266.8-188.1t\\\\274.022t=5782.33\rightarrow t=21.1^oC

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How many half-lives are required for the concentration of reactant to decrease to 1.56% of its original value?4247.56.56
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Answer:

6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.

Explanation:

Using integrated rate law for first order kinetics as:

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Where,

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given:

Concentration is decreased to 1.56 % which means that 0.0156 of [A_0] is decomposed. So,

\frac {[A_t]}{[A_0]} = 0.0156

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\frac {[A_t]}{[A_0]}=e^{-k\times t}

0.0156=e^{-k\times t}

kt = 4.1604

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k=\frac {ln\ 2}{t_{1/2}}

\frac{4.1604}{t}=\frac {ln\ 2}{t_{1/2}}

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<u>6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.</u>

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