Jocelyn has 400 pounds of chemical A that she must react with 200 pounds of chemical B to create chemical C for use in an indust
2 answers:
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Answer:
The correct answers are:
- Krypton: density= 2.8 g/L
- Molar Mass= 63.99 g/mol
- Mass of O₂= 15.29 g
Explanation:
The general equation of an ideal gas is the folllowing:
P x V = n x R x T
Where: P= pressure (in atm), V= volume; n= number of moles, R= gas constant (0,082 L.atm/K.mol) and T= temperature (in K).
<u>For krypton</u>:
P= 671 mmHg = 0,882 atm
V= 478 ml x 1000 ml/1 L= 0,478 L
T= 47ºC= 320 K
MM= 83.8 g/mol (from Periodic Table, Kr is an inert gas so it is a monoatomic gas)
P x V = n x R x T
Since the number of moles of a compound can be calculated by dividing the mass of compound (m) into its molar mass (MM):
n= m/MM
We can replace the expression in the first equation to obtain:
m/V=
Density (d) is equal to the mass per volume (m/V), so we can directly calculate the density:
d= m/V= \frac{P x MM}{R x T}=
= (0.882 atm x 83.8 g/mol)/(0.082 L.atm/K.mol x 320 K)
= 2.81 g/L
<u>For the gas:</u>
d= 2.18 g/L
T= 66ºC= 339 K
P= 720 mmHg= 0.947 atm
d= \frac{P x MM}{R x T}
⇒MM =
= (2.18 g/L x 0.082 L.atm/K.mol x 339 K)/(0.947 atm)
= 63.99 g/mol ≅ 64 g/mol
<u>For the O₂</u>:
V= 5.60 L
P= 1.75 atm
T= 250 K
MM(O₂) = 2 x Atomic Mass O= 2 x 16 g/mol= 32 g/mol
We can use the second equation:
P x V= \frac{m}{MM} x R x T
⇒ m = = (1.75 atm x 5.6 L x 32 g/mol)/(0.082 L.atm/K.mol x 250 K)
= 15.29 g≅ 16 g
Answer:
One complete revolution around a circular path.
Explanation:
Let us take the case of a car moving in a circular track of radius r metres.
In one revolution, the car covers the length(distance) equal to the perimeter of the circle.
In this case, distance traveled = 2r metres
But after one complete revolution, the car reaches the same position as it was at the beginning of the motion.
Hence, the initial and final points coincide or the car hasn't changed it's position w.r.t the initial point.
So in this case, the displacement is zero.
Hence, revolution of a car around a circular path is an example of an object traveling a distance but having no displacement.