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.
Suspensions
Explanation:
Suspensions are heterogeneous mixtures that contains large particles that can settle out or be filtered.
- Suspensions are mixtures of small insoluble particles of a solid in a liquid or gas.
- Examples are:
- powdered chalk in water
- muddy water
- harmattan
The particles in suspension can settle on standing
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heterogeneous mixture brainly.com/question/1446244
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Answer:
D
Explanation:
Gibb's free energy change(∆G) and Standard electrode potential of electrochemical (Ecell) determine the spontaneity of a reaction.
when ∆G > 0, the reaction is not spontaneous
∆G < 0, the reaction is spontaneous
∆G = 0, the reaction is in equilibrium
when Ecell > 0, the redox reaction is spontaneous
Ecell < 0, the redox reaction is not spontaneous
Ecell = 0, the redox reaction is in equilibrium.
The answer Fam is B) Models
volume of H₂O = 7.2 L
Explanation:
The combustion reaction of methane (CH₄):
CH₄ + 2 O₂ → CO₂ + 2 H₂O
Now we calculate the number of moles of methane using the following formula:
number of moles = volume / 22.4 (L/mole)
number of moles of CH₄ = 3.6 / 22.4
number of moles of CH₄ = 0.16 moles
Taking in account the chemical reaction, we devise the following reasoning:
if 1 mole of CH₄ produce 2 moles of H₂O
then 0.16 moles of CH₄ produce X moles of H₂O
X = (0.16 × 2) / 1 = 0.32 moles of H₂O
And now we can calculate the volume of water (H₂O) produced by the reaction:
number of moles = volume / 22.4 (L/mole)
volume = number of moles × 22.4 (L/mole)
volume of H₂O = 0.32 × 22.4
volume of H₂O = 7.2 L
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combustion reaction
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