Question

Please help me with it

What was the weight percent of water in the hydrate before heating? Put answer in one decimal place. weight of water O.505 weight of hydrate 2.209 How many moles of water were lost if the amount of water lost was 0.415 grams? Do not include units and assume three significant figures m all numbers. Be sure to include the zero before the decimal if the number is less than one. How many moles of water are there for each mole of salt in the following formula? answer in numerical format. MgSO_4 middot 1 H_2O If the crucible originally weighs 3.715 g and 2.000 g of hydrate are added to it. what is the weight of the water that is lost if the final weight of the crucible and anhydrous salt s 5.168? Do not include units and assume three decimal places for all numbers. Remember to include the zero before the decimal if the number is less than one.

Answer 1

2. Weight percentage of water before heating = (weight of water/weight of hydrate)*100 = (0.505/2.029)*100 = 24.88%

3. NUmber of moles of water = weight of H2O/mol. wt of H2O = 0.415/18 = 0.023 moles

4. The given formula contain 1mole of H2O for eacch mole of formula of compound

5. Weight of crucible = 3.715 g

Weight of hydrate added = 2.000 g

Total weight = 3.715+2.000 = 5.715 g

Weight of crucible+anhydrous salt = 5.168

Weight of water = 5.715 + 5.168 = 0.547 g