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2 years ago

You would need these tools to set up a filtration device

Chemistry

1 answer:

marysya [2.9K]2 years ago

3 0

1/2 L of swamp water (or tap water with mud or dirt added)

2 L soda pop bottle with its lid

2 L plastic soda pop bottle—cut in half (have an adult help with cutting)

1000 ml beaker

2 20 oz cups

1 tbsp alum (aluminum potassium sulfate)

2 c fine sand

1 c coarse sand

1 c small pebbles

1 filter paper or coffee filter

1 rubber band

1 large spoon for stirring

spoon for scooping the alum

stopwatch or timer

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HELP!!! PLEASE!! I don't know this one question!!!

lara31 [8.8K]

Lead(II) Sulfate, the second answer

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3 years ago

What does chemistry have to do with drug testing

diamong [38]

Answer:

Forensic drug chemists analyze samples of unknown materials including powders, liquids and stains to determine the chemical identity or characteristics of the compounds that make up the sample. samples submitted as evidence in a drug-related case can contain one compound or a mixture of many compounds.

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3 years ago

An ethylene glycol solution contains 21.4 g of ethylene glycol (C2H6O2) in 97.6 mL of water. (Assume a density of 1.00 g/mL for

8090 [49]

Answer: The freezing point and boiling point of the solution are $-6.6^0C$ and $101.8^0C$ respectively.

Explanation:

Depression in freezing point:

$T_f^0-T^f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_f$ = freezing point of solution = ?

$T^o_f$ = freezing point of water = $0^0C$

$k_f$ = freezing point constant of water = $1.86^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(0-T_f)^0C=1\times (1.86^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_f=-6.6^0C$

Therefore,the freezing point of the solution is $-6.6^0C$

Elevation in boiling point :

$T_b-T^b^0=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_b$ = boiling point of solution = ?

$T^o_b$ = boiling point of water = $100^0C$

$k_b$ = boiling point constant of water = $0.52^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(T_b-100)^0C=1\times (0.52^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_b=101.8^0C$

Thus the boiling point of the solution is $101.8^0C$

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3 years ago

Plz help ASAP. thx for you answers ALGEBRA 2 sorry

iren [92.7K]

So, the answer to 27.) would be <em>2x.</em> Both 6x and 2x can be divided by 2x, but they can't go any higher without the end-answer becoming a fraction. As such, 2x is the greatest common factor.

For 28.), x and x^2 can't be like terms, since like terms have the same variable and exponent :)

Hope I could help!

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3 years ago

In which way does a na1+ ion differ from a neutral na atom?

DochEvi [55]

Na 1+ misses 1 electron

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3 years ago

You would need these tools to set up a filtration device​

You would need these tools to set up a filtration device