This is similar to the experiment you are doing in lab though the answer may be

Question

This is similar to the experiment you are doing in lab though the answer may be different. A 3.50 g sample of Epsom salt was heated for five minutes and the resulting solid, MgSO_4, had a mass of 1.85 g. From this data determine the number of moles of water lost and the amount of MgSO_4, that remains. Using your data, what is the value of x in the formula for Epsom salt, MgSO_4 middot xH_2O? Explain the following observations using arguments that includes Intermolecular Forces of Attraction. a) Iodine, I_2, dissolves in hexane, C_6H_14, but not in water. b) Water boils at 100 degree C but H_2S boils at -60 degree C. c) In the series of molecules from CH_4 to CCl_4 boiling point increases as the number of chlorine atoms increases:

Explanation / Answer

Question 5.

m = 3.5 g of Esom salt

solid afterwards --> MgSO4

m = 1.85 g of MgSO4

find numbe of moles of water lost...

A)

first, find mass of water after heating:

mass of water = mass of sample - mass of dry salt = 3.50-1.85 = 1.65 g of water

moles of water = mass/MW = 1.65/18 = 0.091666 moles of water

moles of dry salt = mass/MW = 1.85/120.366 = 0.01536

find MgSO4: H2O ratio

ratio = water moles / dry salt moles = 0.091666 /0.01536 = 5.9678 --> nearest integer 6

so

MgSO4 * 6 H2O

Q6.

a)

they are both, nonpolar, therefore, like dissolves like, water is polar so it wil lnever mix with I2 or C6H14. The london dispersion forces will not allow it

b)

Water will form Hydrogen bonding, so H ------ O bond adda energy required to "break" this and change form liquid to vapor phase

c)

Clearly, boiling must increase drastically as we increase Chlorine atoms, since it is a heavy atom

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