The compounds that are produced upon this combustion reaction would be Carbon Dioxide and water.
CO2 = Carbon and Oxygen
H2O = Hydrogen and Oxygen.
The exact molecular amounts or moles can be determined by balancing this combustion reaction.
<u>Answer:</u> The value of 
for the net reaction is 
<u>Explanation:</u>
The given chemical equations follows:
<u>Equation 1:</u> 
<u>Equation 2:</u> 
The net equation follows:
As, the net reaction is the result of the addition of first equation and the second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the second equilibrium constant.
The value of equilibrium constant for net reaction is:
We are given:
Putting values in above equation, we get:
Hence, the value of for the net reaction is
Answer: 6.75 moles
Explanation:
This is a simple stoichiometry proboe. that I would set up like this:
(13.5 moles CuCI2) (1 mol I2 / 2 moles CuCi2)
That means you all you have to do for this problem is divide by 2 and cancel out the unit moles CuCI2, which leaves you with 6.75 moles I2.
Hope this helps :)
Answer:
See explanation
Explanation:
In the Rutherford experiment, alpha particles were directed at the same spot on a thin gold foil.
As the alpha particles hit the foil, most of the alpha particles went through the foil. In Rutherford's interpretation, most of the particles went through because the atom consisted largely of empty space.
However, some of the alpha particles were deflected through large angles, in Rutherford's interpretation, the deflected alpha particles had hit the dense positive core of the atom which he called the nucleus.
This accounted for their scattering through large angles throughout the foil in all directions.
V(NaOH)=15 mL =0.015 L
C(NaOH)=0.1 mol/L
C(H₂SO₄)=0.05 mol/L
2NaOH + H₂SO₄ = Na₂SO₄ + 2H₂O
n(NaOH)=V(NaOH)C(NaOH)=2n(H₂SO₄)
n(H₂SO₄)=V(H₂SO₄)C(H₂SO₄)
V(NaOH)C(NaOH)=2V(H₂SO₄)C(H₂SO₄)
V(H₂SO₄)=V(NaOH)C(NaOH)/{2C(H₂SO₄)}
V(H₂SO₄)=0.015*0.1/{2*0.05}=0.015 L = 15 mL
