For the excited state of Ca at the absorption of 422.7 nm light,the energy difference is mathematically given as
E= 4.70x10-22 kJ/mol
<h3>What is the energy difference (kJ/mole) between the ground and the first excited state?</h3>
Generally, the equation for the Energy is mathematically given as
E = nhc / λ
Where
h= plank's constant
h= 6.625x 10-34 Js
c = speed of light
c= 3x 108 m/s
Therefore
E = 1*(6.625x 10-34 Js)( 3x 10^8 m/s) / ( 422.7x10^-9)
E= 4.70x10-22 kJ/mol
In conclusion, Energy
E= 4.70x10-22 kJ/mol
Read more about Energy
brainly.com/question/13439286
Answer:
The answer is 1.15m.
Since molality is defined as moles of solute divided by kg of solvent, we need to calculated the moles of H2SO4 and the mass of the solvent, which I presume is water.
We can find the number of H2SO4 moles by using its molarity
C=nV→nH2SO4=C⋅VH2SO4=6.00molesL⋅48.0⋅10−3L=0.288
Since water has a density of 1.00kgL, the mass of solvent is
m=ρ⋅Vwater=1.00kgL⋅0.250L=0.250 kg
Therefore, molality is
m=nmass.solvent=0.288moles0.250kg=1.15m
Answer:
<h2>73.53 mL</h2>
Explanation:
The volume of a substance when given the density and mass can be found by using the formula
From the question we have
We have the final answer as
<h3>73.53 mL</h3>
Hope this helps you
Mass of methanol (CH3OH) = 1.922 g
Change in Temperature (t) = 4.20°C
Heat capacity of the bomb plus water = 10.4 KJ/oC
The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change.
Let’s assume that no heat is lost to the surroundings. First, let’s calculate the heat changes in the calorimeter. This is calculated using the formula shown below:
qcal = Ccalt
Where, qcal = heat of reaction
Ccal = heat capacity of calorimeter
t = change in temperature of the sample
Now, let’s calculate qcal:
qcal = (10.4 kJ/°C)(4.20°C)
= 43.68 kJ
Always qsys = qcal + qrxn = 0,
qrxn = -43.68 kJ
The heat change of the reaction is - 43.68 kJ which is the heat released by the combustion of 1.922 g of CH3OH. Therefore, the conversion factor is:
It is the electron sharing.
electronegative element + electronegative element
exemple :
O₂ , H₂
hope this helps!
