Frequency of photon = is 6.85 X 10¹⁴ sec⁻¹
Energy of photon , E = hv
where h is Planck's constant, v is the frequency of photon
E = 6.63 × 10⁻³⁴ J.s x 6.85 X 10¹⁴ s⁻¹
E = 4.54 x 10⁻¹⁹ J
Therefore, the energy of a photon of green light is 4.54 x 10⁻¹⁹ J.
This is equivalent to having a standard enthalpy change of reaction equal to 10.611 kJ
<u>Explanation</u>:
The standard enthalpy change of reaction, Δ
H
∘
, is given to you in kilojoules per mole, which means that it corresponds to the formation of one mole of carbon dioxide.
C
(s] + O
2(g]
→
CO
2(g]
Remember, a negative enthalpy change of reaction tells you that heat is being given off, i.e. the reaction is exothermic.
First to convert grams of carbon into moles,
use carbon's molar mass(12.011 g).
Moles of C = mass in gram / molar mass
= 0.327 g / 12.011 g
Moles of C = 0.027 moles
Now, in order to determine how much heat is released by burning of 0.027 moles of carbon to form carbon-dioxide.
= 0.027 moles C
393 kJ
Heat released = 10.611 kJ.
So, when 0.027 moles of carbon react with enough oxygen gas, the reaction will give off 10.611 kJ of heat.
This is equivalent to having a standard enthalpy change of reaction equal to 10.611 kJ
Potassium and calcium fluoride are both metals
Answer:
5.52atm
Explanation:
Using the pressure law formula:
P1/T1 = P2/T2
Where;
P1 = initial pressure (atm)
P2 = final pressure (atm)
T1 = initial temperature (K)
T2 = final temperature (K)
According to the question, the following information were provided;
P1 = 4.72 atm
P2 = ?
T1 = -3.50°C = -3.50 + 273 = 269.5K
T2 = 42°C = 42 + 273 = 315K
Using P1/T1 = P2/T2
4.72/269.5 = P2/315
CROSS MULTIPLY
4.72 × 315 = 269.5 × P2
1,486.8 = 269.5P2
P2 = 1,486.8 ÷ 269.5
P2 = 5.52atm
Solution :
A cell that is concentrated is constructed by the same half reaction for the anode as well as he cathode.
We know,
In a standard cell,
the reduction half cell reaction is :

The oxidation half ell reaction :

Thus the complete reaction of the cell is :

cell = 