Answer:
V=0.68L
Explanation:
For this question we can use
V1/T1 = V2/T2
where
V1 (initial volume )= 0.75 L
T1 (initial temperature in Kelvin)= 303.15
V2( final volume)= ?
T2 (final temperature in Kelvin)= 273.15
Now we must rearrange the equation to make V2 the subject
V2= (V1/T1) ×T2
V2=(0.75/303.15) ×273.15
V2=0.67577931717
V2= 0.68L
Answer:
9.474 x 10^2
Explanation:
ok. first you have to get the value in the required unit so 9474mm/(10mm/cm) = 947.4 so scientific notation states that the number must be raised to any power of an integer and the value of the number being raised must be less than than 10 and more than or equal to 1
so it must have one digit in front so.. 947.4 becomes 9.474 and because you move 2 places to the left, ur power is positive 2
and proof 10^2 is 100 so multiply 9.474 by 100 and u will get 947.4 cm which is also 9474 mm
V₁ = initial Volume of the balloon after it is blown up = 365 L
V₂ = new Volume of the balloon after it is taken outside = ?
T₁ = initial temperature of the balloon = 283 K
T₂ = new temperature of the balloon = 300 K
using the equation
V₁/V₂ = T₁/T₂
365/V₂ = 283/300
V₂ = 387 L
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.
