Answer:
The chemist would require to use 43.43 grams.
Explanation:
In order to solve this problem we need to know<u> how much do 0.550 moles of selenium weigh</u>. To do that we use selenium's<em> molar mass </em>and multiply it by the given number of moles:
- 0.550 mol * 78.96 g/mol = 43.43 g
The chemist would require to use 43.43 grams.
Answer:
The electrons are lost from the valence shell (outermost electron shell) of the atom.
Explanation:
This is able to be inferred not only because valence electrons being lost first is a trend but also because the atom in question has actually 3 valence electrons (13-2-8 = 3).
Answer:
0.17 lb
Explanation:
78 g * (1 lb/454 g)=0.17 lb
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Answer : The amount of oxygen gas collected are, 0.217 mol
Explanation :
Using ideal gas equation :
where,
P = pressure of gas = 
(1 atm = 760 torr)
V = volume of gas = 5 L
T = temperature of gas = 
n = number of moles of gas = ?
R = gas constant = 0.0821 L.atm/mol.K
Now put all the given values in the ideal gas equation, we get:
Thus, the amount of oxygen gas collected are, 0.217 mol
