1)How many kilograms of magnesium can be obtained from 5.96 km^3 of seawater if

Question

1)How many kilograms of magnesium can be obtained from 5.96 km^3 of seawater if the initial Mg^2+ concentration is .19% by mass. Remember the density of seawater is 1.04 g/ml

2) how many grams of MgCl2 must dissolve to equal the ion concentration of 62.3 grams of NaCl? also how many grams of CaS must dissolve?

3) find the ph of two equivalence points and the volume (ml) of .407 M HNO3 needed to reach them in the titration of .188 L of .250 M ethylenediamine(H2NCH2CH2NH2)

first equiv point___ ph____ 2nd equiv point___ ph___

4)what volumes of .200 M HCOOH and 2.0 M NaOH would make .450 L of a buffer with the same ph as a buffer made from 475 ml of .2 M benzoic acid and 25 ml of 2.0 M NaOH?

Explanation / Answer

volume of sea water = 5.96 km^3 = 5.96*10^9 m^3 = 5.96*10^15 ml

mass of sea water = 5.96*10^15 * 1.04 = 6.19*10^15 gm

mass of mg^+2 = 0.19 *6.19*10^13 = 1.177 *10^13 g = 1.177*10^10 kg

=> 1.177*10^10 kg

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