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:
C. Lithium is most easily oxidized of the metals listed on the activity series and therefore it will most easily give electrons to metal cations
Explanation:
"Lithium" is a type of alkali metal that has a "single valence electron." Since it is a reactive element, it easily gives up an electron when it is combined with other elements. Such giving up of electron is meant to create compounds or bonds.
Among the common metals listed, "lithium" is the most easily oxidized. This means that it donates its electrons immediately. Such combination makes it exist as a<em> "cation"</em> or <em>"positively-charged."</em>
So, this explains the answer.
4 carbon electrons. Hope this helps have a nice day
Mass =70 ( Mass of protons=1 ,Mass of neutrons =1, Mass of electron =0.0005(can be ignored))
Therefore, 40 +30=70
Charge= -2 ( it is taking in/attracting electrons to its shell) base on the proton number you are able to identify if it is attracting or releasing an electron, if the electron number is more than proton number then it is attracting therefore resulting in a negative charge vice versa for releasing an electron.
Answer:
0,07448M of phosphate buffer
Explanation:
sodium monohydrogenphosphate (Na₂HP) and sodium dihydrogenphosphate (NaH₂P) react with HCl thus:
Na₂HP + HCl ⇄ NaH₂P + NaCl <em>(1)</em>
NaH₂P + HCl ⇄ H₃P + NaCl <em>(2)</em>
The first endpoint is due the reaction (1), When all phosphate buffer is as NaH₂P form, begins the second reaction. That means that the second endpoint is due the total concentration of phosphate that is obtained thus:
0,01862L of HCl×
= 1,862x10⁻³moles of HCl ≡ moles of phosphate buffer.
The concentration is:
= <em>0,07448M of phosphate buffer</em>
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I hope it helps!
