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Ray Of Light [21]
3 years ago
5

Could volume by displacement be used to determine the volume of a lump of rock salt? Explain would be nice

Chemistry
1 answer:
Readme [11.4K]3 years ago
5 0

Yes it could, but you'd have to set up the process very carefully.
I see two major challenges right away:

1).  Displacement of water would not be a wise method, since rock salt
is soluble (dissolves) in water.  So as soon as you start lowering it into
your graduated cylinder full of water, its volume would immediately start
to decrease.  If you lowered it slowly enough, you might even measure
a volume close to zero, and when you pulled the string back out of the
water, there might be nothing left on the end of it.

So you would have to choose some other fluid besides water ... one in
which rock salt doesn't dissolve.  I don't know right now what that could
be.  You'd have to shop around and find one.

2).  Whatever fluid you did choose, it would also have to be less dense
than rock salt.  If it's more dense, then the rock salt just floats in it, and
never goes all the way under.  If that happens, then you have a tough
time measuring the total volume of the lump.

So the displacement method could perhaps be used, in principle, but
it would not be easy.


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Help with the LAB answers
galben [10]

Answer : The reaction rate at 3^oC,24^oC,40^oC\text{ and }65^oC are 36mg/L/sec,142mg/L/sec,190mg/L/sec\text{ and }352mg/L/sec

Solution : Given,

Mass of tablet = 1000 mg

Volume of water = 0.200 L

Formula used :

\text{Reaction rate}=\frac{\text{Mass of tablet}/\text{Volume of water}}{\text{Reaction time}}

Now we have to calculate the reaction rate at different temperatures and reaction time.

\text{Reaction rate at }3^oC=\frac{1000mg/0.200L}{138.5sec}=\frac{5000mg/L}{138.5sec}=36mg/L/sec

\text{Reaction rate at }24^oC=\frac{1000mg/0.200L}{34.2sec}=\frac{5000mg/L}{34.2sec}=142mg/L/sec

\text{Reaction rate at }40^oC=\frac{1000mg/0.200L}{26.3sec}=\frac{5000mg/L}{26.3sec}=190mg/L/sec

\text{Reaction rate at }65^oC=\frac{1000mg/0.200L}{14.2sec}=\frac{5000mg/L}{14.2sec}=352mg/L/sec

Therefore, the reaction rate at 3^oC,24^oC,40^oC\text{ and }65^oC are 36mg/L/sec,142mg/L/sec,190mg/L/sec\text{ and }352mg/L/sec

5 0
4 years ago
Read 2 more answers
How much energy is involved to cool 16.2 g of steam at 115.0°C to -2.00°C?
love history [14]

Explanation:

think its specific heat of water = 4.18 J/g°Ci

6 0
3 years ago
How many signals would you expect to find in the 1h nmr spectrum of 1 butanol?
guajiro [1.7K]
The ¹H-NMR of 1-Butanol is shown below,

There are Five signals found in the proton NMR spectrum of 1-Butanol.

A Broad Singlet signal at ≈ 4.7 ppm (Down Field = More Deshielded) is for proton directly attached to Oxygen atom.

Triplet Signal at ≈ 3.65 ppm (Down Field = Deshielded) is for methylene group directly attached to Oxygen atom.

Multiplet Signal at ≈ 1.6 ppm is for methylene which is present between two methylene groups. The shielding effect is increasing as moving away from Oxygen atom.

Multiplet Signal at ≈ 1.4 ppm is for methylene which is present between terminal methyl and methylene groups. The shielding effect is increasing as moving away from Oxygen atom.

Triplet Signal at ≈ 0.9 ppm (Up Field = Most Shielded) is for terminal methyl group attached to methylene group.

Result:
           Five Signals are found in the ¹H-NMR of 1-Butanol.

4 0
3 years ago
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