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laiz [17]
3 years ago
15

PLEASE HELP!?!.!,! What are the products of the combustion of a hydrocarbon?

Chemistry
2 answers:
kicyunya [14]3 years ago
8 1

Answer:

That would be B. Carbon Dioxide and Water

Explanation:

I literally just took the unit test like a day ago on edge and got it right, but give the other guy brainliest because he is smort.

Tatiana [17]3 years ago
3 0

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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Ok, got it!!! thank you so much
bekas [8.4K]
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7 0
2 years ago
Write balanced nuclear equations for the following:(a) β⁻ decay of silicon-32
S_A_V [24]

The balanced nuclear equations for the following:(a) β⁻ decay of silicon-32 is (27,14)Si -> (0,-1)beta + (27,15)P

<h3>What is balanced nuclear equation?</h3>

A nuclear reaction is generally expressed by a nuclear equation, which has the general form, where T is the target nucleus, B is the bombarding particle, R is the residual product nucleus, and E is the ejected particle, and Ai and Zi (where I = 1, 2, 3, 4) are the mass number and atomic number, respectively. Finding a well balanced equation is critical for understanding nuclear reactions. Balanced nuclear equations provide excellent information about the energy released in nuclear reactions. Balancing the nuclear equation requires equating the total atomic number as well as the total mass number before and after the reaction using the rules of atomic number and mass number conservation in a nuclear reaction.

To learn more about nuclear equations visit:

brainly.com/question/12221598

#SPJ4

7 0
1 year ago
"11. Barium nitrate reacts with aqueous sodium sulfate to produce solid barium sulfate and aqueous sodium nitrate. Abigail place
Amanda [17]

Answer:

44 mL of Na2SO4

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

Ba(NO3)2 (aq) + Na2SO4 (aq) —> BaSO4 (s) + 2NaNO3 (aq)

Step 2:

Determination of the number of mole of Ba(NO3)2 in 20.00 mL of 0.500 M barium nitrate (Ba(NO3)2). This is illustrated below:

Molarity of Ba(NO3)2 = 0.5 M

Volume of solution = 20 mL = 20/1000 = 0.02 L

Mole of solute (Ba(NO3)2) =?

Molarity = mole /Volume

0.5 = Mole of Ba(NO3)2 / 0.02

Cross multiply to express in linear form

Mole of Ba(NO3)2 = 0.5 x 0.02

Mole of Ba(NO3)2 = 0.01 mole

Step 3:

Determination of the number of mole of Na2SO4 that reacted.

Ba(NO3)2 (aq) + Na2SO4 (aq) —> BaSO4 (s) + 2NaNO3 (aq)

From the balanced equation above,

1 mole of Ba(NO3)2 reacted with 1 mole of Na2SO4.

Therefore, 0.01 mole of Ba(NO3)2 will also react with 0.01 mole of Na2SO4.

Step 4:

Determination of the volume of Na2SO4 needed for the reaction. This is illustrated below:

Mole of Na2SO4 = 0.01 mole

Molarity of Na2SO4 = 0.225M

Volume =?

Molarity = mole /Volume

0.225 = 0.01 / volume

Cross multiply to express in linear form

0.225 x Volume = 0.01

Divide both side by 0.225

Volume = 0.01/0.225

Volume of Na2SO4 = 0.044 L

Converting 0.044 L to mL, we have

Volume of Na2SO4 = 0.044 x 1000

Volume of Na2SO4 = 44 mL

Therefore, 44 mL of Na2SO4 is needed for the reaction

6 0
3 years ago
Read 2 more answers
The solubility product for an insoluble salt with the formula M2X3 is written as ________, where s is the molar solubility
masha68 [24]

Answer:

36s^5

Explanation:

We have;

M2X3 (s)------> 2M^3+(aq)  + 3X^2-(aq)

If [M^3+(aq)] = [X^2-(aq)] = s

We then have;

Ksp = (2s)^2 * (3s)^3

Ksp = 4s^2 * 9s^3

Ksp = 36s^5

Note that Ksp is known as the solubility product. It is an equilibrum equation that shows the solubility of a solute in water.

3 0
2 years ago
How many moles of AgCl2 are in 5.78x1024 molecules of AgCl2?
Zolol [24]

Answer:

9.6 mol AgCl2

Explanation:

You have to use Avogadro's number: 6.023 x 10^23

5.78 x 10^24 molecules (1 mol AgCl2/ 6.023 x 10^23 molecules) =9.6 mol AgCl2

3 0
3 years ago
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