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2 years ago

Balance the combustion reaction between butane and oxygen. 2C4H10 + O2 → CO2 + H2O

Chemistry

2 answers:

dimaraw [331]2 years ago

6 0

<h2>Answer:</h2><h3>Balanced equation:</h3>

2C4H10 + 8O2 → 8CO2 + 10H2O

<h3>Description:</h3>

There should be 8 molecules of oxygen which react with 2 molecules of butane to produce 8 molecules of carbon dioxide and 10 molecules of water.

pashok25 [27]2 years ago

4 0

Answer:

Explanation:

Answer:

2C4H10 + 13O2 → 8CO2 + 10H2O

Explanation:

2C4H10 + O2 → CO2 + H2O

There are 8 carbon atoms on the left side of the equation thus , there should be equal number of carbon atoms on the right side

So, if we balance carbon atoms only, the new eqation becomes

2C4H10 + O2 → 8CO2 + H2O

Now carbon atoms are balance but there are 20 , hydrogen atoms on the left side but 2 atoms only on the right side. So the new equation after balancing hydrogen atom becomes

2C4H10 + O2 → 8CO2 + 10H2O

Now both carbon and hydrogen atoms are balanced but oxygen atoms are pending. There are 26 atoms of (O) on the right side , while 1 molecule of (O2)

So the balanced equation would be

2C4H10 + 13O2 → 8CO2 + 10H2O

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kotegsom [21]

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3 years ago

Read 2 more answers

What was the eutectic temperature (temperature from the two lines of best fit cross) for the mixture

Ymorist [56]

Answer:

hello your question is incomplete below is the missing part of the question

answer : 104°c

Explanation:

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2 years ago

Arrange the following H atom electron transitions in order of decreasing wavelength of the photon absorbed or emitted:

Katyanochek1 [597]

The decreasing order of wavelengths of the photons emitted or absorbed by the H atom is : b → c → a → d

Rydberg's formula :

$\frac{1}{\lambda} = R_h (\frac{1}{n_1^2}-\frac{1}{n_2^2} )$ ,

where λ is the wavelength of the photon emitted or absorbed from an H atom electron transition from $n_1$ to $n_2$ and $R_h$ = 109677 is the Rydberg Constant. Here $n_1$ and $n_2$ represents the transitions.

(a) $n_1$ =2 to $n_2$ = infinity

$\frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{\infty^2} )$ = 109677/4 [since 1/infinity = 0] Therefore, $\lambda$ = 4 / 109677 = 0.00003647 m