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Alenkasestr [34]
3 years ago
12

What is the mass of 1.63x10^21

Chemistry
1 answer:
krok68 [10]3 years ago
3 0

Answer:

7.60 x 10^-2 g.

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PLEASE HELP!?!.!,! What are the products of the combustion of a hydrocarbon?
Tatiana [17]

Answer:

carbon dioxide and water

Explanation:

Example: Combustion of Methane (CH₄(g))

CH₄(g) + 2O₂(g) => CO₂(g) + 2H₂O(g)**

____________________

Note: The combustion of any hydrocarbon produces CO₂ & H₂O. That is,

Ethane (C₂H₆) + O₂ => CO₂(g) + H₂O(g)

Propane (C₃H₈) + O₂ => CO₂(g) + H₂O(g)

Butane (C₄H₁₀) + O₂ => CO₂(g) + H₂O(g)

The issue remaining is to balance the reaction equation. For these type equation balance Carbon 1st, then Hydrogen and finish with Oxygen. Balancing in this order leaves Oxygen which can be balanced using fractions. If problem requires lowest whole number ratios of elements, simply multiply entire equation by 2 to get standard equation*

______________________

*Standard Equation is defined as the smallest whole number ratios of elements. The 'standard equation' is significant in that it is assumed to be at STP conditions; i.e., 0⁰C (=273K) & 1.0 Atmosphere pressure.

  • Ethane (C₂H₆) + 7/2O₂(g) => 2CO₂(g) + 3H₂O(g)

    => 2C₂H₆ + 7O₂(g) => 4CO₂(g) + 6H₂O(g)  <= Standard Form of Rxn

  • Propane (C₃H₈) + 5O₂(g) => 3CO₂(g) + 4H₂O(g) <= Standard Form of Rxn (no need to balance with the '2' multiple)
  • Butane (C₄H₁₀) + 13/2O₂ => 4CO₂(g) + 5H₂O(g)

    => 2C₃H₈ + 13O₂(g) => 4CO₂(g) + 5H₂O(g) <= Standard Form of Rxn

______________________

**Also, note that water, H₂O(g), is listed as a gas. In some cases it will be listed as a liquid, H₂O(l).

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4 years ago
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The Correct Answer Is 3.2
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3 years ago
A small piece of hot metal is placed in cooler water. The metal is left in the water
scoundrel [369]

Answer: The amount of energy lost by the metal is equal to the amount of energy gained by the water

Explanation:

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What is 0.738 × 104? Enter your answer in the box.<br> _____
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The answer would be 76.752 .
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Animal cells don't have a dividing cell wall like plant cells do

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