Answer:
8o/58*6*10power23 gives u the answer. molecules=moles*avagadrono
moles=weight/molecular wt
hope u understandmark me as branliest
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: See below
Explanation:
1. a) 0.15 moles calcium carbonate (15g/100g/mole)
b) 0.15 moles CaO (molar ratio of CaO to CaCO3 is 1:1)
c) 8.4 grams CaO (0.15 moles)*(56 grams/mole)
2. a) 0.274 moles Na2O (17g/62 grams/mole)
b) 46.6 grams NaNO3 (2 moles NaNO3/1 mole Na2O)*(0.274 moles Na2O)*(85 g/mole NaNO3)
Answer:
C: The actual yield depends on the reaction conditions, but the theoretical yield varies only with reactant amounts
Explanation:
Looking at the options, the correct one is Option C because the actual yield usually depends on the conditions of the reaction, while the theoretical yield usually varies with only the amount of reactant.
