Use the data table to answer the questions below mass of
evaporating .07g mass of
evaporating dish plus .12 g mass of
evaporating dish and anhydrous crystals .58g
if the hydrate used in the lab for the above data table is
mgso4.7h20, then calculate the theoretical % water in the hydrate
15%
51%
73%
95%
100%

Option B is correct : 51%

Explanation:

Data Given:

Hydrates used in lab = MgSO₄ .7H₂O

% water in hydrate = ?

Solution:

As we look for the the theoretical % water in hydrates

So

For H₂O atomic masses are given below

H = 1 g/mol

O = 16 g/mol

Then find the total mass of H₂O in hydrated salt

           7H₂O = 7 (2x1 +1x16) g/mol

           7H₂O = 7 (2+16) g/mol

           7H₂O = 7 (18) g/mol

Now,

> find total Molar Mass of Molecule:

       Molar Mass of MgSO₄ • 7H₂O = [24 + 32 + 16x4] + 7(2+16)

       Molar Mass of MgSO₄ • 7H₂O = [24+ 32 + 64] + 126

       Molar Mass of MgSO₄ • 7H₂O = [120] + 126

Now to find the mass by percent of (water) H₂O

Formula used to find the mass by percent of a component

Percent composition of H₂O = mass of H₂O in Molecule / molar mass of MgSO₄•7H₂O x 100%

Put the values in above formula

        Percent by mass of H₂O = 126 (g/mol) / 246 (g/mol) x 100%

        Percent by mass of H₂O = 0.51 x 100%

        Percent by mass of H₂O = 51%

So,

Option B is correct : 51%

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be