<span>Determine the root-mean-square sped of CO2 molecules that have an average Kinetic Energy of 4.21x10^-21 J per molecule. Write your answer to 3 sig figs.
</span><span>
E = 1/2 m v^2
If you substitute into this formula, you will get out the root-mean-square speed.
If energy is Joules, the mass should be in kg, and the speed will be in m/s.
1 mol of CO2 is 44.0 g, or 4.40 x 10^1 g or 4.40 x 10^-2 kg.
If you divide this by Avagadro's constant, you will get the average mass of a CO2 molecule.
4.40 x 10^-2 kg / 6.02 x 10^23 = 7.31 x 10^-26 kg
So, if E = 1/2 mv^2
</span>v^2 = 2E/m = 2 (4.21x10^-21 J)/7.31 x 10^-26 kg = 115184.68
Take the square root of that, and you get the answer 339 m/s.
The correct answer is 7 I just took the test :)
Nitrogen fixation is the process that makes atmospheric nitrogen available to plants by mutualistic and free-living bacteria. The process is undertaken by the rhizobium bacteria that live in root roots of plants such as legumes. The mutualistic relationship is that the plant supplies the bacteria with a habitat in which to live, water, and nutrients, and the bacteria supply nitrogen for making plant proteins.
Nrjrjffjjdrjrjrjejejdjrjrjrjejrrjdjdkdkrjjrjrjrrjrjrjrjrjrjrjrjrkrk
Explanation:
The given data is as follows.
Pressure (P) = 760 torr = 1 atm
Volume (V) = 
= 0.720 L
Temperature (T) = 
= (25 + 273) K = 298 K
Using ideal gas equation, we will calculate the number of moles as follows.
PV = nRT
Total atoms present (n) = 
= 
= 0.0294 mol
Let us assume that there are x mol of Ar and y mol of Xe.
Hence, total number of moles will be as follows.
x + y = 0.0294
Also, 40x + 131y = 2.966
x = 0.0097 mol
y = (0.0294 - 0.0097)
= 0.0197 mol
Therefore, mole fraction will be calculated as follows.
Mol fraction of Xe = 
= 
= 0.67
Therefore, the mole fraction of Xe is 0.67.
