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.
Answer:
Homogeneous
Explanation:
Homogeneous mixtures are uniform in composition. They have the same proportion of components throughout. Homogeneous mixtures are called solutions. Sugar, paint, alcohol, gold are all examples of homogeneous mixtures because they look the same throughout.
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Answer: 72.93 litres
Explanation:
Given that:
Volume of gas (V) = ?
Temperature (T) = 24.0°C
Convert 24.0°C to Kelvin by adding 273
(24.0°C + 273 = 297K)
Pressure (P) = 1.003 atm
Number of moles (n) = 3 moles
Molar gas constant (R) is a constant with a value of 0.0821 atm L K-1 mol-1
Then, apply ideal gas equation
pV = nRT
1.003 atm x V = 3.00 moles x 0.0821 atm L K-1 mol-1 x 297K
1.003 atm•V = 73.15 atm•L
Divide both sides by 1.003 atm
1.003 atm•V/1.003 atm = 73.15 atm•L/1.003 atm
V = 72.93 L
Thus, the volume of the gas is 72.93 litres
Answer:
5.41 ×10⁻²²
Explanation:
We were told right from the question that both the Zinc ions and the Zinc oxide adopts a face-centered cubic arrangement.
Then, the number of ZnO molecule in one unit cell = 4
The standard molar mass of ZnO = 81.38g
Avogadro's constant = 6.023 × 10²³ mole
∴
The mass of one unit cell of zinc oxide can be calculated as:
= 
= 5.40461564×10⁻²²
≅ 5.41 ×10⁻²²
∴ The mass of one unit cell of zinc oxide = 5.41 ×10⁻²²
