Answer:
C shake it repeatedly
Explanation:
i think thats the answer if not sorry hope it helps you.
<span>From the Born–Landé equation the lattice energy U ∝ (Z+ × Z-) / (r+ + r-) where Z+ and Z- are the charges on the cation and anion, respectively; r+ and r- are radii of the cation and anion, respectively.
The Z+×Z- term dominates. MgO has 2×2 (4) so it will have the higher U than Li2O Z+ × Z- = 2.
You only consider (r+ + r-) term when the Z+×Z- term is the same; smallest the (r+ + r-), the larger is the U.
U(MgO) = 3795 </span>kJ mol^-1
<span>U(Li2O) = 2799 kJ mol^-1
MgO has larger.
</span>
Answer: Theoretical yield is 313.6 g and the percent yield is, 91.8%
Explanation:
To calculate the moles :
According to stoichiometry :
1 mole of 
require 3 moles of 
Thus 2.8 moles of will require=
of
Thus is the limiting reagent as it limits the formation of product and is the excess reagent.
As 1 mole of give = 2 moles of 
Thus 2.8 moles of give =
of
Mass of 
Theoretical yield of liquid iron = 313.6 g
Experimental yield = 288 g
Now we have to calculate the percent yield
Therefore, the percent yield is, 91.8%
*The molality of a solution is calculated by taking the moles of solute and dividing by
the kilograms of solvent* Basically if we had 1.00 mole of sucrose (it's about 342 3 grams) and
proceeded to mix it into exactly 1.00 liter water. It would dissolve and make sugar
water. We keep adding water, dissolving and stirring until all the solid was gone. We
then made sure everything was well-mixed.
What would be the molality of this solution? Notice that my one liter of water weighs
1000 grams (density of water = 1.00 g / mL and 1000 mL of water in a liter).
Mixture is a combination of two or more substances. The components of the mixture are not chemically reacted and can be separated using physical means.
compound is a combination of two or more substances that are chemically reacted or bound together. The components of the compound cannot be separated except by chemical reactions.
Answer:
<span>(4) consist of two or more substances</span>
