The answer is a because a fill in the blank is always just one word
Answer:
86.3 g of N₂ are in the room
Explanation:
First of all we need the pressure from the N₂ in order to apply the Ideal Gases Law and determine, the moles of gas that are contained in the room.
We apply the mole fraction:
Mole fraction N₂ = N₂ pressure / Total pressure
0.78 . 1 atm = 0.78 atm → N₂ pressure
Room temperature → 20°C → 20°C + 273 = 293K
Let's replace data: 0.78 atm . 95L = n . 0.082 . 293K
(0.78 atm . 95L) /0.082 . 293K = n
3.08 moles = n
Let's convert the moles to mass → 3.08 mol . 28g /1mol = 86.3 g
Answer:
F centripetal force (tension) = 275.9 N
Explanation:
Given data:
Mass = 1.50 kg
Radius = 0.520 m
Velocity of ball = 9.78 m/s
Tension = ?
Solution:
F centripetal force (tension) = m.v² / R
F centripetal force (tension) = 1.50 kg . (9.78 m/s)² / 0.520 m
F centripetal force (tension) = 1.50 kg . 95.65 m²/s² / 0.520 m
F centripetal force (tension) = 143.5 kg. m²/s² / 0.520 m
F centripetal force (tension) = 275.9 N
Answer:
his is an example of a first-year chemistry question where you must first convert two of the pressures to the units of the third and add them up, per Dalton’s law of additive pressures. There are three possible answers, one for each of the three pressure units.
1 atm = 760 torr …… torr and mm Hg are the same
1 atm = 101.3 kPa
Dalton’s law:
P(total) = P(O2) + P(N2) + P(CO2)
Explanation:
Gases will assume whatever pressure depending on the equation of state of the mixture (in this case) and the volume htey are contained in. That could be the ideal gas law and simple mixing law, If you are quoting the partial pressures which you call simply “the pressure” of each gas, and that these refer to their values in the present mixture, then yes, we would add them up. The pressures are low enough for the ideal gas law to apply provided the temperature is not extremely low as well .
hey mate here is ur answer
solution
mass{m}=3 gram
=3/1000
volume{v}=16cm
=16/100
density=m/v
=3/1000÷16/100
=3/160
=0.01875kg/m3
