Answer:
41.17g
Explanation:
We are given the following parameters for Flourine gas(F2).
Volume = 5.00L
Pressure = 4.00× 10³mmHG
Temperature =23°c
The formula we would be applying is Ideal gas law
PV = nRT
Step 1
We find the number of moles of Flourine gas present.
T = 23°C
Converting to Kelvin
= °C + 273k
= 23°C + 273k
= 296k
V = Volume = 5.00L
R = 0.08206L.atm/mol.K
P = Pressure (in atm)
In the question, the pressure is given as 4.00 × 10³mmHg
Converting to atm(atmosphere)
1 mmHg = 0.00131579atm
4.00 × 10³ =
Cross Multiply
4.00 × 10³ × 0.00131579atm
= 5.263159 atm
The formula for number of moles =
n = PV/RT
n = 5.263159 atm × 5.00L/0.08206L.atm/mol.K × 296K
n = 1.0834112811moles
Step 2
We calculate the mass of Flourine gas
The molar mass of Flourine gas =
F2 = 19 × 2
= 38 g/mol
Mass of Flourine gas = Molar mass of Flourine gas × No of moles
Mass = 38g/mol × 1.0834112811moles
41.169628682grams
Approximately = 41.17 grams.
Answer : Electron P has greater energy difference than the Electron N.
Explanation :
Wavelength range of violet light = 400 - 500 nm
Wavelength range of orange light = 600 - 700 nm
The Planck's equation is,
where,
E = energy of light
c = speed of light
= wavelength of light
According to the Planck's equation, wavelength and energy follow inverse relation. As the wavelength increases, energy decreases.
From the given spectrum, the wavelength of violet light is less. We conclude that When electron P gives violet light on transition it means that energy difference between the energy level was high.
From the given spectrum, the wavelength of orange light is more. We conclude that When electron N gives orange light on transition it means that energy difference between the energy level was low.
So, Electron P which gives violet light on transition has greater energy difference than the Electron N.
Heat the solution till all the water molecules are no longer in the solution.
Or add more solute till it stops dissolving.(maximum point of dissolvtion.)
Answer:
–36 KJ.
Explanation:
The equation for the reaction is given below:
2B + C —› D + E. ΔH = – 24 KJ
From the equation above,
1 mole of D required – 24 KJ of energy.
Now, we shall determine the energy change associated with 1.5 moles of D.
This can be obtained as illustrated below:
From the equation above,
1 mole of D required – 24 KJ of energy
Therefore,
1.5 moles of D will require = 1.5 × – 24 = –36 KJ.
Therefore, –36 KJ of energy is associated with 1.5 moles of D.
