A substance that has no specific volume changes to a substance that has a specific volume.
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.
Answer:
Mass = 141.6 g
Explanation:
Given data:
Mass of Kr in gram = ?
Volume in L = 9.59 L
Temperature = 46.0°C
Pressure = 4.62 atm
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
Now we will convert the temperature.
46.0+273 = 319 K
4.62 atm × 9.59 L = n× 0.0821 atm.L/ mol.K ×319 K
44.3 atm.L = n×26.19 atm.L/ mol
n = 44.3 atm.L / 26.19 atm.L/ mol
n = 1.69 mol
Mass in gram:
Mass = number of moles × molar mass
Mass = 1.69 mol × 83.79 g/mol
Mass = 141.6 g
<span>1 mole glucose gives 2 moles of ethanol
moles of glucose in 2.4 kg = 2400 / 180.18 = 13.320 moles
so moles of ethanol produced = 2* 13.32 = 26.64 moles
weight of ethanol 26.64 * 46.07
=1227.30 gm or 1.23 Kg</span>