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) CuCl2

b) CH3CH2CH3

c) CS2

d) Cr

e) CH3CH3

2007-03-06 14:37:47 · 5 answers · asked by Jamgirl108 1 in Science & Mathematics Chemistry

5 answers

I myself would go with what you have as A.) because B, C, and E are organic i.e. non-polar and will not dissolve in water.

A is a salt which they tend to like water.

2007-03-06 14:47:51 · answer #1 · answered by brian_holinsworth1 2 · 0 0

Ones that will dissolve and why
CuCl2 : I have a table on the back of my periodic table which tells me the solublilty of different substances. Have a look in your book to see if you have this table.

One that won't dissolve and why
CH3CH2CH3 : This is an alkane and alkanes are insoluble in water.
CH3CH3 : This is an alkane also.
Cr - It is a metal and isn't in ionic form.
CS2 - 0.2 g/100 ML of water (20 °C). I calculated that there is 2.62x10^-2mol per Litre. A substance is insoluble if less than 0.01 mol of the substance dissolves per litre.

2007-03-06 23:01:59 · answer #2 · answered by Tigeress 2 · 0 0

a) CuCl2 - Copper Chloride is a salt will dissolve
b) CH3CH2CH3 - is an olefin and likely a gas
c) CS2 - Carbonyl Sulfide and too tightly bonded
d) Cr - Chrome is a metal no way
e) CH3CH3 - ethane is a hydrocarbon gas is soluble but not dissolvable in water

2007-03-06 22:45:46 · answer #3 · answered by lostlatinlover 3 · 0 0

Jamba - lia

When you mean water, you mean tapp water? Distilled water (H to the 2O).... clarify and the reason and what you seek will be come apparent.......


Prediction:
tomorrow in Chemistry class you will fall asleep, you will awake up with a wee bit of slobber on your lip glossed lips, no one will notice, you will dream about other things you learned that day to ask on here, the responder to your next question is not a potential marriage candidate.

The power of the haiku has been unleashed on you, use it wisely.

Dissolve in water
Melt away, strong are ions
H minus rips it

An ode to acidic H-

2007-03-06 22:50:48 · answer #4 · answered by Anonymous · 0 0

b ch3ch2ch3

2007-03-06 22:42:11 · answer #5 · answered by Anonymous · 0 1

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