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IRINA_888 [86]
3 years ago
13

Balance equation C5H10+O2=CO2+H2O

Chemistry
2 answers:
Crazy boy [7]3 years ago
7 0

Answer:

2 C5H10 + 15 O2 => 10 CO2 + 10 H2O

Explanation: Each individual atom both in the reactant and product side have equal number of atoms.

Reactant Product

Carbon 10 10

Hydrogen 20 20

Oxygen 30 30

Oksanka [162]3 years ago
4 0

Answer:

2C5H10 + 15O2 —> 10CO2 + 10H2O

Explanation:

C5H10 + O2 —> CO2 + H2O

To balance the equation, do the following:

There are 5 atoms of C on the left side of the equation and 1atom on the right side. It can be balance by putting 5 in front of CO2 as shown below:

C5H10 + O2 —> 5CO2 + H2O

There are 10 atoms of H on the left side and 2 atoms on the right. It can be balance by putting 5 in front of H2O as shown below:

C5H10 + O2 —> 5CO2 + 5H2O

Now, there are a total of 15atoms of O2 on the right side and 2 atoms on the left side. It can be balance by putting 15/2 in front of O2 as shown below:

C5H10 + 15/2O2 —> 5CO2 + 5H2O

Now multiply through by 2 to clear the fraction as shown below:

2C5H10 + 15O2 —> 10CO2 + 10H2O

Now the equation is balanced

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Calculate the percent errorin a length measurementof 4.45cm if the correct value is 4.06
FinnZ [79.3K]

Answer:

9.6 %

Explanation:

<u>Step 1: How to define  percent error ? </u>

⇒ % error is the difference between a measured value and the known or accepted value

⇒Percent error is calculated using the following formula:

⇒%error  =  | Experimental value-theoretical value/theoretical value |  x100%

⇔ this can be written as well as : error = (| Experimental value/ theoretical value | - | Theoretical value / Theoretical value | ) x100%

<u>Step 2: Calculate % error</u>

In this case, this means :

%error = ( |(4.45 cm - 4.06cm ) / 4.06cm | ) x100%

%error = 0.096 x100%

%error =9.6 %

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3 years ago
Be<br><br> Name the element.<br><br> Number of shells?<br><br> Valence electrons?
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Answer:

Name the element: Beryllium

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Valence electrons: 2

Explanation:

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3 years ago
NEED ASAP
solmaris [256]
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It isn't symmetrical.
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Hydrogen sulfide

Explanation:

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The reaction is shown below:

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A general reaction is also shown below of the reaction of Iron(II) sulfide with any nonoxidizing acid, HX as:

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