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kogti [31]
2 years ago
13

HELPP PLEASE!!!WILL PROB GIVE BRAINLY!!! could someone please show a visual representation of 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O

involved in a reaction
Chemistry
1 answer:
jeka942 years ago
3 0

Question:

2C

8

H

18

+ 25O

2

 

→

16CO

2

+ 18H

2

O

The combustion of octane, C

8

H

18

, proceeds according to the reaction above. If 346 mol of octane combusts, what volume of carbon dioxide is produced at 37.0 degrees Celsius and 0.995 atm?

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What is the empirical formula of an oxide of nitrogen containing 63.61% by mass of nitrogen and 36.69% by mass of oxygen?
katen-ka-za [31]

Answer:

D. N₂O

Explanation:

Let's assume we have 100 g of the compound.  That means it consists of 63.61 grams of nitrogen and 36.69 grams of oxygen.

Converting masses to moles:

63.61 g N × (1 mol N / 14.01 g N) = 4.540 mol N

36.69 g O × (1 mol O / 16.00 g O) = 2.293 mol O

Normalize by dividing by the smallest:

4.540 / 2.293 = 1.980 mol N

2.293 / 2.293 = 1.000 mol O

So there is approximately twice as many N atoms as O atoms.  The empirical formula is therefore N₂O.

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Jet001 [13]

Answer:

CaS, CaBr₂, VBr₅, and V₂S₅.

Explanation:

  • The ionic compound should be neutral; the overall charge of it is equal to zero.
  • Binary ionic compound is composed of two different ions.

<u>Ca²⁺ can combined with either Br⁻ or S²⁻ to form binary ionic compounds.</u>

  • CaS can be formed via combining Ca²⁺ with S²⁻ to form the neutral binary ionic compound CaS.
  • CaBr₂ can be formed via combining 1 mole of Ca²⁺ with 2 moles of Br⁻ to form the neutral binary ionic compound CaBr₂.

<u>V⁵⁺ can combined with either Br⁻ or S²⁻ to form binary ionic compounds.</u>

  • V₂S₅ can be formed via combining 2 moles of V⁵⁺ with 5 moles of S²⁻ to form the neutral binary ionic compound V₂S₅.
  • VBr₅ can be formed via combining 1 mole of V⁵⁺ with 5 moles of Br⁻ to form the neutral binary ionic compound VBr₅.

<em>So, the empirical formula of four binary ionic compounds that could be formed is: CaS, CaBr₂, VBr₅, and V₂S₅.</em>

<em></em>

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drek231 [11]

Answer:

Explanation:

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