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: Th enthalpy of combustion for the given reaction is 594.244 kJ/mol
Explanation: Enthalpy of combustion is defined as the decomposition of a substance in the presence of oxygen gas.
W are given a chemical reaction:



To calculate the enthalpy change, we use the formula:

This is the amount of energy released when 0.1326 grams of sample was burned.
So, energy released when 1 gram of sample was burned is = 
Energy 1 mole of magnesium is being combusted, so to calculate the energy released when 1 mole of magnesium ( that is 24 g/mol of magnesium) is being combusted will be:

Answer:
113.69°k
Explanation:
V1=85L of helium V2=32L
T1= 29°C +273= 302°K T2=?
T2=<u>TIV2</u>
V1
T2=<u>(302)(32)</u>= <u>9664</u>
85 85
T2= 113.69°K
Answer:
See explanation and image attached
Explanation:
The standard cell potential at 298 K is given by;
E°cathode - E°anode
Hence;
E°cell = 0.34 V - (-0.76 V)
E°cell = 0.34 V + 0.76 V
E°cell = 1.1 V
To reduce Zn^2+ to Zn then Zn must be the cathode, hence;
E°cell = (-0.76 V) - 0.34 V
E°cell = -1.1 V
They are lost from valence shell (it's outermost shell of an atom).
->>> outside the atom