Answer:
Mg would blow off. AI would be affective to copper but not to MG
Explanation:
Mass = mass/molar mass of ch3ch2nh2
the molar mass of CH3CH2NH2 = 12 +(1x3)+12+(1 x2)+14+(1x2) =45 g/mol
moles is therefore = 1.50g /45g/mol = 0.033 moles
Answer:
127.0665 amu
Explanation:
Firstly, to answer the question correctly, we need to access the percentage compositions of the iodine and the contaminant iodine. We can do this by placing their individual masses over the total and multiplying by 100%.
We do this as follows. Since the mass of the contaminant iodine is 1.00070g, the mass of the 129I in that particular sample will be 12.3849 - 1.00070 = 11.3842g
The percentage abundances is as follows:
Synthetic radioisotope % = 1.0007/12.3849 * 100% = 8.1%
Since there are only two constituents, the percentage abundance of the 129I would be 100 - 8.1 = 91.9%
Now, we can use these percentages to get the apparent atomic mass. We get this by multiplying the percentage abundance’s by the atomic masses of both and adding together.
That is :
[8.1/100 * 128.9050] + [91.9/100 * 126.9045] = 10.441305 + 116.6252355 = 127.0665 amu
<span>The answer is 4. The molecules of each material entice each other over dispersion (London) intermolecular forces. Whether a substance is a solid, liquid, or gas hinge on the stability between the kinetic energies of the molecules and their intermolecular magnetisms. In fluorine, the electrons are firmly apprehended to the nuclei. The electrons have slight accidental to stroll to one side of the molecule, so the London dispersion powers are comparatively weak. As we go from fluorine to iodine, the electrons are far from the nuclei so the electron exhausts can more effortlessly misrepresent. The London dispersion forces developed to be increasingly stronger.</span>
