32.8 g of Butane is required and 99.3 g of CO₂ is produced
<u>Explanation:</u>
The above mentioned reaction can be written as,
C₄H₁₀(g) + 13 O₂(g) → 4CO₂(g) + 5 H₂O(g) where ΔH (rxn)= -2658 kJ
It is given that 1.5 × 10³ kJ of energy is produced, the original reaction says that 2658 kJ of heat is produced, which means that less than one mole of butane is used in the reaction.
That is
of butane reacted
Now this moles is converted into mass by multiplying it with its molar mass = 0.564 mol × 58.122 g / mol
= 32.8 g of butane.
Mass of CO₂ produced = 0.564 ×44.01 g /mol × 4 mol
= 99.3 g of CO₂
Thus 32.8 g of Butane is required and 99.3 g of CO₂ is produced
Answer:
189.2 KJ
Explanation:
Data Given
wavelength of the light = 632.8 nm
Convert nm to m
1 nm = 1 x 10⁻⁹
632.8 nm = 632.8 x 1 x 10⁻⁹ = 6.328 x 10⁻⁷m
Energy of 1 mole of photon = ?
Solution
Formula used
E = hc/λ
where
E = energy of photon
h = Planck's Constant
Planck's Constant = 6.626 x 10⁻³⁴ Js
c = speed of light
speed of light = 3 × 10⁸ ms⁻¹
λ = wavelength of light
Put values in above equation
E = hc/λ
E = 6.626 x 10⁻³⁴ Js ( 3 × 10⁸ ms⁻¹ / 6.328 x 10⁻⁷m)
E = 6.626 x 10⁻³⁴ Js (4.741 x 10¹⁴s⁻¹)
E = 3.141 x 10⁻¹⁹J
3.141 x 10⁻¹⁹J is energy for one photon
Now we have to find energy of 1 mole of photon
As we know that
1 mole consists of 6.022 x10²³ numbers of photons
So,
Energy for one mole photons = 3.141 x 10⁻¹⁹J x 6.022 x10²³
Energy for one mole photons = 1.89 x 10⁵ J
Now convert J to KJ
1000 J = 1 KJ
1.89 x 10⁵ J = 1.89 x 10⁵ /1000 = 189.2 KJ
So,
energy of one mole of photons = 189.2 KJ
Answer:
Carbon, Hydrogen, Oxygen, Nitrogen, and Sulfur.
(Those are all the ones I know)
Answer:
B. NUCLEAR POWER CAN PRODUCE ELECTRICITY IN A Y KIND IF WEATHER.
