Answer:
131 atm
Explanation:
To find the new pressure, you need to use Boyle's Law:
P₁V₁ = P₂V₂
In this equation, "P₁" and "V₁" represent the initial pressure and volume. "P₂" and "V₂" represent the final pressure and volume. You can find the new pressure (P₂) by plugging the given values into equation and simplifying.
P₁ = 3.88 atm P₂ = ? atm
V₁ = 7.74 L V₂ = 0.23 L
P₁V₁ = P₂V₂ <----- Boyle's Law
(3.88 atm)(7.74 L) = P₂(0.23 L) <----- Insert values
30.0312 = P₂(0.23 L) <----- Simplify left side
131 = P₂ <----- Divide both sides by 0.23
Answer:
0.6258 g
Explanation:
To determine the number grams of aluminum in the above reaction;
- determine the number of moles of HCl
- determine the mole ratio,
- use the mole ratio to calculate the number of moles of aluminum.
- use RFM of Aluminum to determine the grams required.
<u>Moles </u><u>of </u><u>HCl</u>
35 mL of 2.0 M HCl
2 moles of HCl is contained in 1000 mL
x moles of HCl is contained in 35 mL
We have 0.07 moles of HCl.
<u>Mole </u><u>ratio</u>
6HCl(aq) + 2Al(s) --> 2AlCl3(aq) + 3H2(g)
Hence mole ratio = 6 : 2 (HCl : Al
- but moles of HCl is 0.07, therefore the moles of Al;
Therefore we have 0.0233333 moles of aluminum.
<u>Grams of </u><u>Aluminum</u>
We use the formula;
The RFM (Relative formula mass) of aluminum is 26.982g/mol.
Substitute values into the formula;
The number of grams of aluminum required to react with HCl is 0.6258 g.
The answer for this question would be B) False or the second option.
Answer is FALSE: ✅
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:
have you tried c
Explanation:
the chicken and I don't know if you can make it
