Answer:
17.934 kg of water
Explanation:
If balanced equation is not given; this format can come in handy.
For any alkane of the type : CₙH₂ₙ₊₂ , it's combustion reaction will follow:
2CₙH₂ₙ₊₂ + (3n+1) O₂ → (2n)CO₂ + 2(n+1) H₂O
For butane:
2C₄H₁₀(g) + 13O₂(g) → 8CO₂(g) + 10H₂O(l)
2 moles of butane gives 10 moles of water.
1 mol of any substance has Avogadro number(N) of molecules in it( 6.022 x 10²³)
Mass of 1 mole of any substance is equal to it's molar mass
So, if 2 x N molecules of butane gives 10 x 18 g of water.
Then 1.2 x 10²⁶ molecules will give:
= 17.934 x 10³ g of water
= 17.934 kg of water
Kinetic energy is energy that comes from motion. Anything that is currently in motion has kinetic energy.
Let’s look at each example to determine if they have kinetic energy.
First off, a car in the garage: let’s ask ourselves- Is the car in motion?
No, it is sitting in the garage. It is not moving; therefore it doesn’t have any kinetic energy.
Next, a box sitting on a shelf: let’s ask ourselves the same question- Is the box in motion?
No, it is sitting on the shelf. Again, it is not moving. It doesn’t have any kinetic energy.
Our third item is a ball lodged in a tree: again, we will ask ourselves the same question- Is the object moving?
No, it isn’t moving. Again, since it is not moving, it will not have kinetic energy.
Our last item is a frisbee flying through the air: asking ourselves the same question- Is it moving?
Yes, the object is moving. Yes, it has kinetic energy.
The frisbee flying through the air has kinetic energy.
D. dormancy is the correct answer
Answer:
Standard free-energy change at 
is 
Explanation:
Oxidation: 
Reduction: 
--------------------------------------------------------------------------------------
Overall: 
Standard cell potential, 
So, 
We know, standard free energy change at (
): 
where, n is number of electron exchanged during cell reaction, 1F equal to 96500 C/mol
Here n = 2
So, 
Answer:
412.1kJ
Explanation:
For the reaction , from the question -
4Fe (s) + 3O₂ (g) → 2Fe₂O₃ (s)
Δ Hrxn = Δ H°f (products) - Δ H°f (reactants)
In case the compound is in its standard state , enthalphy of formation is zero
Hence ,
for the above reaction ,
ΔHrxn =( 2 * Δ H° (Fe₂O₃ )) - [ ( 4 *Δ H° Fe ) + (3 * Δ H° O₂ )]
The value for Δ H°(Fe₂O₃ ) = - 824.2kJ/mol
Δ H° Fe = 0
Δ H° O₂ = 0
Putting in the above equation ,
ΔH rxn = ( 2 * Δ H° (Fe₂O₃ )) - 0
ΔHrxn = 2× - 824.2 kJ / mol = - 1648.4 kJ/mol
- 1648.4 kJ/mol , this much heat is released by the buring of 4 mol of Fe.
Hence ,
for 1 mol of Fe ,
- 1648.4 kJ/mol / 4 = 412.1kJ
