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Classify substances as either pure substances or mixtures (heterogeneous or homogeneous)

Submitted by Anonymous (not verified) on Mon, 07/28/2008 - 17:55

Hey, so i've been doing my summer chemistry homework assignment and it is proving to be rather difficult.

So I had to classify the substances as either pure substances or mixtures and if it was a mixture hetero. or homo., this is what I got:

a) concrete - pure substance
b) sea water- homo. mixture
c) magnesium- pure substance
d) gasoline- homo. mixture
e)air- homo mixture
f) tomato juice- homo mixture
g) iodine crystal- pure substance
e) A nickel- pure substance

I'm not sure on a few of these answers and feedback would be really helpful =]=]

ANOTHER QUESTION:

A solid white substance A is heated strongly in the abssence of air. It decomposes to form a new white solid substance B and a gas C. THe gas has exactly the same properties as the product obtained when carbon is burned with excess oxygen. What can you say whether solids A and B and the gas C are elements or compounds?

I put that:

A: compound
B:??????
C: compound

NEXT QUESTION:

These are calculations problems that I have no idea what to do though i've tried various techniques yet i have been getting varying answers.

1. Calculate the mass of 1.00 ft^3 of gold metal.
my final answer was 588.87 grams but I am very unsure.

2. Calculate the mass of 12.0 cm sphere of lead metal.
I am really confused on how to do this question.

3. THIS ONE IS SOOO HARD!!!!
A 200.00mg piece of gold can be hammered into a sheet that is 2.4 ft by 10ft. a) what is the thickness of this sheet in m?? (b) quote the answer using the SI prefix so that there is no exponent needed.

THANK YOU FOR ANY HELP YOU GIVE!!!


a) concrete - pure substance  No - it is a heterogeneous mixture of cement, sand, and gravel

tomato juice- homo mixture  No, it is a heterogeneous mixture - homogeneous solutions are clear.  Tomato juice is cloudy, thus it must be heterogeneous.

e) A nickel- pure substance  No, it is an alloy of Ni and Cu.  Alloys are homogeneous mixtures because they are uniform throughout.

2nd Question:  The original compound must have been a carbonate (A).  Carbonates decompose to yield metal oxides (a compound-B) plus carbon dioxide (a compound-C).

1. Calculate the mass of 1.00 ft^3 of gold metal.
my final answer was 588.87 grams but I am very unsure.

1.00 cubic ft is a volume, to determine the mass you need to multiply the volume by the density.  The only problem is getting units to cancel:

            1.00 ft3  x    ( 12 in/1 ft )3    x  (100 cm / 39.4 in)3 = 2.82 x 104 cm3  Volume of Gold metal in cm3

                      2.82 x 104 cm3  x  19.3 g/cm3  =  5.45 x 105 g  =  54.5 kg of Au

2. Calculate the mass of 12.0 cm sphere of lead metal.
I am really confused on how to do this question.
I think that we can assume that 12.0 cm sphere is referring to the diameter of the sphere, which means its radius is 6 cm.  This problem is similar to the last problem in that volume x density = mass.  So all we have to do is find the volume, look up the density of Pb and multiply.

So, we need to know that the volume of any sphere is given by the following equation:
                    Volume = (4/3 ) x  pi  x r3

The value of pi is 3.14 and the radius is 6 cm, so we can plug in and find the volume.  I'll let you look up the volume and do the calculation.

3. THIS ONE IS SOOO HARD!!!!
A 200.00mg piece of gold can be hammered into a sheet that is 2.4 ft by 10ft. a) what is the thickness of this sheet in m?? (b) quote the answer using the SI prefix so that there is no exponent needed.

This is the reverse of the problems you've already done, here you find volume by using the density and the mass:
                      volume  =  mass / density

Once you know the volume (in cm3), you will need to convert your length and width values in feet into cm (as I did above).  Now you make use of the formula for the volume of a rectangular object (which is what the sheet of gold is):
                            volume = length x width  x thickness
Since we know the volume, the length, and the width, we can solve for the thickness.

Finally, convert the thickness in cm into meters.

That's not so hard, is it?