Answer:
1.) AgNO₃
2.) 0.563 moles AgBr
Explanation:
The limiting reagent is the reagent that is used up completely during a reaction. It can be identified by calculating which reactant produces the smallest amount of product. This can be done by determining the number of moles of each reagent (via molarity conversion). and then converting it to moles of the product (via mole-to-mole ratio).
AgNO₃ (aq) + KBr (aq) ---> AgBr (s) + KNO₃ (aq)
Molarity (M) = moles / liters
100 mL = 1 L
AgNO₃
45.0 mL / 100 = 45.0 L
1.25 M = ? moles / 0.450 L
? moles = 0.563 moles
KBr
75.0 mL / 100 = 0.750 L
0.800 M = ? moles / 0.750 L
? moles = 0.600 moles
In this case, there is no need to use the mole-to-mole ratio because all of the coefficients are one in the reaction (the amount of the limiting reagent used is the same amount of product produced). Since AgNO₃ produces the smaller amount of product, it is the limiting reagent.
Explanation:
Calcium is the element that has 2 valence electrons.
Its <span>c.chromatography is the process of separating solutions on the basis of their boiling points </span>
Answer:
There are 10.0 moles of beryllium oxide in a 250 grams sample of the compound.
Explanation:
We can calculate the number of moles (η) of BeO as follows:
![\eta = \frac{m}{M}](https://tex.z-dn.net/?f=%20%5Ceta%20%3D%20%5Cfrac%7Bm%7D%7BM%7D%20)
Where:
m: is the mass = 250 g
M: is the molar mass = 25.0116 g/mol
Hence, the number of moles is:
![\eta = \frac{250 g}{25.0116 g/mol} = 10.0 moles](https://tex.z-dn.net/?f=%20%5Ceta%20%3D%20%5Cfrac%7B250%20g%7D%7B25.0116%20g%2Fmol%7D%20%3D%2010.0%20moles%20)
Therefore, there are 10.0 moles of beryllium oxide in a 250 grams sample of the compound.
I hope it helps you!