The gram-formula mass of Sm is 150.36 u, and the gram-formula mass of O is 15.999 u, so the gram-formula mass of SmO is about 150.36+15.999 = 160.36 g/mol.
So, there are about (9.30 * 10^-3)(160.36)=1.49 grams
Answer: -
After rain fall occurs, water accumulates
.
This water is called ground water. This water slowly flows under the ground through the pore spaces in the soil. The water continues to go under the soil till it reaches a certain point where it can no longer go down.
In this region the water slowly accumulates forming a water reservoir called aquifer. The aquifer is a place under the soil which is saturated with water. The upper surface of this zone of saturation is called the water table.
To calculate the concentration of the base based on the titration, the concept used is the equal of number of equivalence of the acid used to that of the base. From this,
Na x Va = Nb x Vb
For HBr and KOH, molarity is equal to normality. Substituting the known values,
(0.75 M) x (22.6 mL - 0 mL) = Nb x (37.5 mL - 0.5 mL)
Nb = 0.46 N
Mb = 0.46 M
Thus, the concentration of the base is approximately 0.46 M.
