Answer:
Moles of
= 6 moles
Explanation:
The reaction of
and
to make
is:
⇒
The above reaction shows that 2 moles of Sc can react with 3 moles of
to form 
Mole Ratio= 2:3
For 10 moles of Sc we need:
Moles of
= 
Moles of
= 
Moles of
=15 moles
So 15 moles of
are required to react with 10 moles of
but we have 9 moles of
, it means
is limiting reactant.


Moles of ScCl_3= 6 moles
Answer:
185.05 g.
Explanation
Firstly, It is considered as a stichiometry problem.
From the balanced equation: 2LiCl → 2Li + Cl₂
It is clear that the stichiometry shows that 2.0 moles of LiCl is decomposed to give 2.0 moles of Li metal and 1.0 moles of Cl₂, which means that the molar ratio of LiCl : Li is (1.0 : 1.0) ratio.
We must convert the grams of Li metal (30.3 g) to moles (n = mass/atomic mass), atomic mass of Li = 6.941 g/mole.
n = (30.3 g) / (6.941 g/mole) = 4.365 moles.
Now, we can get the number of moles of LiCl that is needed to produce 4.365 moles of Li metal.
Using cross multiplication:
2.0 moles of LiCl → 2.0 moles of Li, from the stichiometry of the balanced equation.
??? moles of LiCl → 4.365 moles of Li.
The number of moles of LiCl that will produce 4.365 moles of Li (30.3 g) is (2.0 x 4.365 / 2.0) = 4.365 moles.
Finally, we should convert the number of moles of LiCl into grams (n = mass/molar mass).
Molar mass of LiCl = 42.394 g/mole.
mass = n x molar mass = (4.365 x 42.394) = 185.05 g.
This is because the relative atomic mass is the sum of the no of protons and neutrons while the atomic no is of the former only. it is not a whole number it is th average weighted masses of the various isotopes of the element multiplied by their corresponding relative abundance. pls thank me and mark me briniest
Temp must be Kelvin
38 C =
<span>
<span>
<span>
311.15
</span>
</span>
</span>
K
Volume at STP = 8.50 liters * (273.15 / 311.15) * (725 / 760) =
<span>
<span>
<span>
7.1182746306
</span>
</span>
</span>
Liters
The formula to use is:
Volume at STP = Present Volume * (273.15 / Present Temp °K) * (Present Pressure (Torr) / 760)
The answer is A) Aluminum Bromide
hope this helps :)