We are given the molar mass of Molybdenum as 95.94 g/mol. Also, the chemical symbol for Molybdenum is Mo. This question is asking for the amount of molecules of molybdenum in a 150.0 g sample. However, since molybdenum is a metal and it is in the form of solid molybdenum, Mo (s), it is not actual a molecule. A molecule has one or more atom bonded together. We will instead be finding the amount of atoms of Molybdenum present in the sample. To do this we use Avogadro's number, which is the amount of atoms/molecules of a substance in 1 mole of that substance.
150.0 g Mo/ 95.94 g/mol = 1.563 moles of Mo
1.563 moles Mo x 6.022 x 10²³ atoms/mole = 9.415 x 10²³ atoms Mo
Therefore, there are 9.415 x 10²³ atoms of Molybdenum in 150.0 g.
Answer:
1. 374.69 K
Explanation:
Hello,
In this case, pure water's boiling point is 373.15 K, thus by considering the boiling point increase equation:
Whereas i=2 since two ionic species are formed,actually, the experimental value is 2.42 so better work with it, thus:
Thus, the required boiling point is:
Regards.
The answer is 2 pairs. The carbon atom has 4 electrons and oxygen atom has 6 electrons. The purpose of shared electron is to get 8 electrons at the outer layer. So the carbon atom needs to share 4 electrons which is 2 pairs with oxygen to be stable.
Gas molecules have more freedom in motion—and gases can be thought of as more “disordered”—than molecules of a solid, which are rigidly held in place. When it comes to phases, the entropy increases as you go from a solid to liquid to gas (the gaseous state having the greatest entropy and the solid state having the least).
So, as a sample of solid iodine sublimes to form gaseous iodine, the entropy of the sample increases.
Answer:
maybe 20 butvam not sure ooo
