570/8.5=67.0 58... you only have to take the natural part, si the answer is 67 students
Solution:
According to the Avogadro's number:
6.022 *10^23 drops per 0.050 g/ drop = 3.011 *10^22 grams per mole of drops
3.011 *10^22 grams per 1 kg / 1000 grams = 3.011 *10^19 kilograms / mole of drops
thus the answers are:
3.0 *10^22 grams per mole of drops
3.0 *10^19 kilograms per mole of drops
And,
In the calculation of how many moles of raindrops in the Pacific Ocean is:
7.08X10^20kg per 3.0 *10^19 kilograms per mole of drops = 23.5 moles of drops
This is the required solution.
Answer:
PNO₂ = 0.49 atm
PN₂O₄ = 0.45 atm
Explanation:
Let's begin with the equation of ideal gas, and derivate from it an equation that involves the density (ρ = m/V).
PV = nRT
n = m/M (m is the mass, and M the molar mass)


PxM = ρRT
ρ = PxM/RT
With the density of the gas mixture, we can calculate the average of molar mass (Mavg), with the constant of the gases R = 0.082 atm.L/mol.K, and T = 16 + 273 = 289 K

0.94Mavg = 63.9846
Mavg = 68.0687 g/mol
The molar mass of N is 14 g/mol and of O is 16 g/mol, than
g/mol and
g/mol. Calling y the molar fraction:

And,


So,





The partial pressure is the molar fraction multiplied by the total pressure so:
PNO₂ = 0.52x0.94 = 0.49 atm
PN₂O₄ = 0.48x0.94 = 0.45 atm
Answer:
Explanation:
Given parameters:
Initial temperature T₁ = 25.2°C = 25.2 + 273 = 298.2K
Initial pressure = P₁ = 0.6atm
Final temperature = 72.4°C = 72.4 + 273 = 345.4K
Unknown:
Final pressure = ?
Solution:
To solve this problem, we use an adaption of the combined gas law where the volume gas is fixed. This simplification results into:

where P and T are temperatures, 1 and 2 are initial and final temperatures.
Input the parameters and solve;
P₂ = 0.7atm
Answer:
177.277amu
Explanation:
the total occuring isotopes for Hafnium is =6.
First isotope had an atomic weight of 173.940amu
Second isotope =175.941amu
Third isotope =176.943amu
Fourth isotope=177.944amu
Fifth isotope. =178.946amu
sixth isotope .179.947amu
<em>Avera</em><em>ge</em><em> </em><em>ato</em><em>mic</em><em> </em><em>wei</em><em>ght</em><em> </em><em>of</em><em> </em><em>Haf</em><em>nium</em><em>=</em><em> </em><em>sum</em><em> </em><em>of</em><em> </em><em>all</em><em> </em><em>the </em><em>atomi</em><em>c</em><em> </em><em>weights</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>iso</em><em>topes</em><em>/</em><em> </em><em>Tota</em><em>l</em><em> </em><em>occu</em><em>ring</em><em> </em><em>isotopes</em>
Thus, 173.940amu+175.941amu+176.943amu+177.944amu+178.946amu+179.947amu.= 1063.661amu
Average atomic weight= 1063.661amu /6 = 177.2768333amu
= 177.277amu to 3 decimal places.