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sukhopar [10]
3 years ago
12

John Dalton believed which of the following about atoms?

Chemistry
1 answer:
8090 [49]3 years ago
3 0
Atoms are real even though they're invisible.
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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).

3 0
3 years ago
Read 2 more answers
calculate the volume occupied by 10g of propane gas, under normal conditions of temperature and pressure
andriy [413]

Answer:

5.5 L

Explanation:

First we <u>convert 10 g of propane gas</u> (C₃H₈) to moles, using its <em>molar mass</em>:

  • 10 g ÷ 44 g/mol = 0.23 mol

Then we <u>use the PV=nRT formula</u>, where:

  • P = 1 atm & T = 293 K (This are normal conditions of T and P)
  • n = 0.23 mol
  • R = 0.082 atm·L·mol⁻¹·K⁻¹
  • V = ?

1 atm * V = 0.23 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 293 K

  • V = 5.5 L
3 0
3 years ago
How to feel peace (&gt;_&lt;)​
snow_lady [41]

Answer:

Pray to ur God to give you some peace

Explanation:

and then listen to some peace full music

I only know that

3 0
2 years ago
Read 2 more answers
1. Compare and contrast the characteristics of metals and nonmetals.
Alona [7]

Answer:

1.Metals

These are very hard except sodium

These are malleable and ductile  pieces

These are shiny

Electropositive in nature

Non-metals

These are soft except diamond

These are brittle and can break down into pieces

These are non-lustrous except iodine

Electronegative in nature

2. The electrochemical series helps to pick out substances that are good oxidizing agents and those which are good reducing agents.In an electrochemical series the species which are placed above hydrogen are more difficult to be reduced and their standard reduction potential values are negative.

3. Arrhenius theory, theory, introduced in 1887 by the Swedish scientist Svante Arrhenius, that acids are substances that dissociate in water to yield electrically charged atoms or molecules, called ions, one of which is a hydrogen ion (H+), and that bases ionize in water to yield hydroxide ions (OH−).

4. The common application of indicators is the detection of end points of titrations. The colour of an indicator alters when the acidity or the oxidizing strength of the solution, or the concentration of a certain chemical species, reaches a critical range of values.

4 0
3 years ago
Read 2 more answers
A cylindrical glass tube of length 27.4cm and radius 4 cm is filled with gas. The empty tube has a mass of 254.3 g. The tube fil
Temka [501]

Density of the gas is 3.05 × 10⁻³ g / cm³.

<u>Explanation:</u>

Volume of the cylinder = π r² h

where r is the radius and h is the height of the height or the length of the glass tube.

Here r = 4 cm and h = 27.4 cm

Volume of the cylinder = 3.14 × 4 × 4 × 27.4 = 1376.6 cm³

We have to find the mass of the gas by subtracting the mass of the tube filled with the substance from the mass of the empty tube.

Mass of the substance = 258.5 - 254.3 = 4.2 g

We have to find the density using the formula as,

$ Density = \frac{mass}{volume}

Plugin the values as,

$ Density = \frac{4.2}{1376.6}

              = 3.05 × 10⁻³ g / cm³

So the Density of the gas is 3.05 × 10⁻³ g / cm³.

3 0
3 years ago
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