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azamat
3 years ago
8

BALANCE THIS EQUATION: C4H10 + O2 = C02 + H20

Chemistry
1 answer:
Lostsunrise [7]3 years ago
5 0

Answer:

{ \rm{C _{4} H_{10}+  \frac{13}{2} O _{2} = 4C0 _{2} + 5H _{2}0}} \\

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What are the answers please help
LenaWriter [7]
Thank you for the free 15 points .
4 0
3 years ago
Is O2 always a double bond?
Dima020 [189]
  • O-O single bonds and H-O single bonds in \text{H}_2\text{O}_2 molecules.
  • H-O single bonds in \text{H}_2\text{O} molecules.
  • O=O double bonds in \text{O}_2 molecules.
<h3>Explanation</h3>

How many valence electrons do atoms in each molecule need for them to be stable?

  • Each H atom needs two valence electrons to be stable.
  • Each atom of an element other than H needs eight valence electron to be stable.
  • There are two H atoms and two O atoms in an  \text{H}_2\text{O}_2 molecule. Atoms in each \text{H}_2\text{O}_2 need 2 \times 1 + 2\times 2 = 6 more electrons to be stable.
  • There are two H atoms and one O atom in an \text{H}_2\text{O} molecule. Atoms in each \text{H}_2\text{O} molecule need 2 \times 1 + 2= 4 more electrons to be stable.
  • There are two O atoms in an \text{O}_2 molecule. Atoms in each \text{O}_2 molecule need 2 \times 2 = 4 more electrons to be stable.

How many chemical bonds in each molecule?

Each chemical bond adds one valence electron to each bonding atom. Each chemical bond connects two atoms. As a result, each chemical bond adds two valence electrons to the molecule.

  • Each \text{H}_2\text{O}_2 molecule needs 6 / 2 = 3 chemical bonds.
  • Each \text{H}_2\text{O} molecule needs 4 / 2 = 2 chemical bonds.
  • Each \text{O}_2 molecule needs 4 /2= 2 chemical bonds.

What chemical bonds are these? Again, each H atom needs only one more valence electron to be stable. It will share only one electron with O and form one H-O bond. The rest of the chemical bonds are between O atoms.

  • There are two H atoms in each \text{H}_2\text{O}_2 molecule, which form two H-O bonds. Two of the three chemical bonds in this molecule are H-O. The other is an O-O single bond between the two O atoms.
  • There are two H atoms in each \text{H}_2\text{O} molecule, which form two H-O bonds. Both chemical bonds in this molecule are H-O. There's no O-O bond in this molecule.
  • There is no H atom in \text{O}_2 molecules. Both chemical bonds are between O atoms. However, there are only two O atoms. There must be two chemical bonds between the two O atoms. That bond will be an O-O double bond.
3 0
3 years ago
Read 2 more answers
As the number of protons increases but the distance between the protons and electron REMAINS THE SAME...
Aloiza [94]

Answer:

The correct option is;

The electronegativity increases

Explanation:

The electronegativity is the measure of an atom's ability to attract a shared electron pair. The electronegativity of an atom is dependent on the atom's atomic number and the separation distance between the electrons in the valence shell and the positively charged nucleus such that an increase in the atomic number results in an increase in electronegativity and an increase in the distance between the valence electrons and the nucleus, leads to a decrease in electronegativity.

8 0
3 years ago
139%
GaryK [48]

<u>Answer:</u>

The percent composition of this compound is 94%

<u>Explanation:</u>

The reaction can be formed as

2 \mathrm{Fe}+3 \mathrm{Cl}_{2} \rightarrow 2 \mathrm{FeCl}_{3}

\frac{\text { Weight of } \mathrm{Cl}_{2}}{\text { 3* Molar Mass of } \mathrm{Cl}_{2}}=\frac{\text { Weight of } \mathrm{Fe}}{2 * \text { Molar Mass of Fe }}

\frac{\text { Weight of } \mathrm{Cl}_{2}}{3 *(2 * 35.5)}=\frac{3.56}{2 * 55.8}

\text { Weight of } C l_{2}=\frac{3.56 * 3 * 71}{2 * 55.8}=6.79 \mathrm{g}

\mathrm{n}\left(\mathrm{Cl}_{2}\right)=\mathrm{m}\left(\mathrm{Cl}_{2}\right) / \mathrm{M}\left(\mathrm{Cl}_{2}\right)=6.79 / 71=0.1 \mathrm{m}

\mathrm{n}(\mathrm{Fe})=\mathrm{m}(\mathrm{Fe}) / \mathrm{M}(\mathrm{Fe})=3.56 / 55.8=0.06 \mathrm{m}

Based on no. of iron reacted,  

\mathrm{n}(\text { moles of } \mathrm{Fe})=\mathrm{n}\left(\text { moles of } \mathrm{FeCl}_{3}\right)

n = m/M

\mathrm{m}\left(\mathrm{FeCl}_{3}\right)=\mathrm{n}^{*} \mathrm{M}=0.06^{*} 162.5=9.75 \mathrm{g}

% composition ofFeCl_3  

=  (9.75 / 10.39)^{*} 100

= 94%

6 0
3 years ago
Can someone plz help me???? will mark as brainliest!!!!!!!!!!!
Nookie1986 [14]

1=c    2=a     3=b

i hope these answer your questions


7 0
3 years ago
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