Answer:
19.264×
atoms are present in 3.2 moles of carbon.
Explanation:
It is known that one mole of each element is composed of Avagadro's number of atoms. This is same for all the elements in the periodic table.
So, as 1 mole of any element = Avagadro's number of atoms = 6.02× atoms
It is as simple as understanding a dozen of anything is equal to 12 pieces of that object.
As here the moles of carbon is given as 3.20 moles, the number of atoms in this mole can be determined as below.
1 mole of carbon = 6.02 × atoms
Then, 3.20 moles of carbon = 3.20 × 6.02 × atoms
Thus, 19.264× atoms are present in 3.2 moles of carbon.
it would need to gain 2 more for it to achieve a stable configuration
the best way to remember this is the noble gasses all have 8 valence electrons and that they are the most stable elements on the periodic table
uhmm, white.
Explanation:
you'll basically look like blind ig
This problem is to use the Claussius-Clapeyron Equation, which is:
ln [p2 / p1] = ΔH/R [1/T2 - 1/T1]
Where p2 and p1 and vapor pressure at estates 2 and 1
ΔH is the enthalpy of vaporization
R is the universal constant of gases = 8.314 J / mol*K
T2 and T1 are the temperatures at the estates 2 and 1.
The normal boiling point => 1 atm (the pressure of the atmosphere at sea level) = 101,325 kPa
Then p2 = 101.325 kPa
T2 = ?
p1 = 54.0 kPa
T1 = 57.8 °C + 273.15K = 330.95 K
ΔH = 33.05 kJ/mol = 33,050 J/mol
=> ln [101.325/54.0] = [ (33,050 J/mol) / (8.314 J/mol*K) ] * [1/x - 1/330.95]
=> 0.629349 = 3975.22 [1/x - 1/330.95] = > 1/x = 0.000157 + 1/330.95 = 0.003179
=> x = 314.6 K => 314.6 - 273.15 = 41.5°C
Answer: 41.5 °C
