Answer:
15912 × 10∧-19 KJ
Explanation:
Given data;
frequency of photon = 2.4 × 10^18 1/s.
Planck's constant = 6.63 × 10∧-34 j.s
Energy = ?
Formula:
E = h × ν
E = 6.63 × 10∧-34 j.s × 2.4 × 10^18 1/s
E= 15.912 × 10∧-16 j
now we will convert the joule into kilo joule,
E = 15.912 × 10∧-16 j /1000 = 15.912 × 10∧-19 KJ
An ecosystem is a large community of living organisms (plants, animals and microbes) in a particular area. The living and physical components are linked together through nutrient cycles and energy flows. Basically where they live
In order to determine the increase in boiling point of a solvent due to the presence of a solute, we use the formula:
ΔT = Kb * m * i
Here, Kb is a property of the solvent, so remains constant regardless of the solute. Moreover, because the concentration m has been fixed, this will also not be considered. In order to determine which solute will have the greatest effect, we must check i, the van't Hoff factor.
Simply stated, i is the number of ions that a substance produces when dissolved. Therefore, the solute producing the most ions will be the one causing the greatest change in boiling point temperature.
Answer:
The calorimeter constant is = 447 J/°C
Explanation:
The heat absorbed or released (Q) by water can be calculated with the following expression:
Q = c × m × ΔT
where,
c is the specific heat
m is the mass
ΔT is the change in temperature
The water that is initially in the calorimeter (w₁) absorbs heat while the water that is added (w₂) later releases heat. The calorimeter also absorbs heat.
The heat absorbed by the calorimeter (Q) can be calculated with the following expression:
Q = C × ΔT
where,
C is the calorimeter constant
The density of water is 1.00 g/mL so 50.0 mL = 50.0 g. The sum of the heat absorbed and the heat released is equal to zero (conservation of energy).
Qabs + Qrel = 0
Qabs = - Qrel
Qcal + Qw₁ = - Qw₂
Qcal = - (Qw₂ + Qw₁)
Ccal . ΔTcal = - (cw . mw₁ . ΔTw₁ + cw . mw₂ . ΔTw₂)
Ccal . (30.31°C - 22.6°C) = - [(4.184 J/g.°C) × 50.0 g × (30.31°C - 22.6°C) + (4.184 J/g.°C) × 50.0 g × (30.31°C - 54.5°C)]
Ccal = 447 J/°C
