Answer:
Explanation:
At constant pressure and temperature, the mole ratio of the gases is equal to their volume ratio (a consequence of Avogadro's law).
Hence, the <em>complete combustion reaction</em> that has a ratio of 100 ml of gaseous hydrocarbon to 300 ml of oxygen, is that whose mole ratio is 1 mol hydrocarbon : 3 mol of oxygen.
Then, you must write the balanced chemical equations for the complete combustion of the four hydrocarbons in the list of choices, and conclude which has such mole ratio (1 mol hydrocarbon : 3 mol oxygen).
A complete combustion reaction of a hydrocarbon is the reaction with oxygen that produces CO₂ and H₂O, along with the release of heat and light.
<u>a. C₂H₄:</u>
- C₂H₄ (g) + 3O₂ (g) → 2CO₂(g) + 2H₂O (g)
Precisely, for this reaction the mole ratio is 1 mol C₂H₄: 2 mol O₂, hence, this is the right choice.
The following analysis just shows that the other options are not right.
<u>b. C₂H₂:</u>
- 2C₂H₂ (g) + 5O₂ (g) → 4CO₂(g) + 2H₂O (g)
The mole ratio for this reaction is 2 mol C₂H₂ :5 mol O₂.
<u>с. С₃Н₈</u>
- C₃H₈ (g) + 5O₂ (g) → 3CO₂(g) + 4H₂O (g)
The mole ratio is 1 mol C₃H₈ : 5 mol O₂
<u>d. C₂H₆</u>
- 2C₂H₆ (g) +7 O₂ (g) → 4CO₂(g) + 6H₂O (g)
The mole ratio is 2 mol C₂H₆ : 7 mol O₂
If magnets are heated to the Curie point, they lose their ability to be magnetic. ... Cooling causes the molecules in the magnet to have less kinetic energy. This means that there is less vibration in the magnet's molecules, allowing the magnetic field they create to be more consistently concentrated in a given direction.
Answer:
c. 9.94 g
Explanation:
From the question,
Using
mt = m₀e⁻kt.................... Equation 1
Where mt = mass of the leaf remaining in the bag, m₀ = original mass of leave that was placed in the bag, k = decay constant, t = time.
Given: m₀ = 33 g, k = 0.04, t = 30 days.
Substitute into equation 1
mt = 33(e⁻(0.04ˣ30))
mt = 33e⁻¹²/¹⁰
mt = 33/e¹²/¹⁰
mt = 33/3.320
mt = 9.94 g.
Hence the right answer is c. 9.94 g
Answer:
i think the answer is helium (top right in periodic table)
The grams of ethane present in a sample containing 0.4271 mole is 12.84 g
<h3>Description of mole </h3>
The mole of a substance is related to it's mass and molar mass according to the following equation
Mole = mass / molar mass
With the above formula, we can obtain the mass of ethane. Details below
<h3>How to determine the mass of ethane</h3>
The following data were obtained from the question:
- Mole of ethane = 0.4271 mole
- Molar mass of ethane = 30.067 g/mol
- Mass of ethane =?
The mass of ethane present in the sample can be obtained as follow:
Mole = mass / molar mass
Cross multiply
Mass = mole × molar mass
Mass of ethane = 0.4271 × 30.067
Mass of ethane = 12.84 g
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