The pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c is 4.1atm.
Therefore, option A is correct option.
Given,
Mass m = 14g
Volume= 3.5L
Temperature T= 75+273 = 348 K
Molar mass of CO = 28g/mol
Universal gas constant R= 0.082057L
Number of moles in 14 g of CO is
n= mass/ molar mass
= 14/28
= 0.5 mol
As we know that
PV= nRT
P × 3.5 = 0.5 × 0.082057 × 348
P × 3.5 = 14.277
P = 14.277/3.5
P = 4.0794 atm
P = 4.1 atm.
Thus we concluded that the pressure will 14. 0 g of co exert in a 3. 5 l container at 75°c is 4.1atm.
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Answer:
At the lowest point in the oscillation, the momentum is zero.
At the lowest point in the oscillation, ![mg < ks_s](https://tex.z-dn.net/?f=mg%20%3C%20ks_s)
Explanation:
Since spring block system is performing to and fro motion along straight line
So here we can say at the lowest position of its path the velocity will become zero.
So we can say that momentum of the spring block system is given as
![P = mv](https://tex.z-dn.net/?f=P%20%3D%20mv)
![P = 0](https://tex.z-dn.net/?f=P%20%3D%200)
Also we know that after reaching the lowest point the block will again go up towards its mean position
So at the lowest point of the spring block system the block will move upwards again
So this will accelerate upwards hence
![F_{spring} > mg](https://tex.z-dn.net/?f=F_%7Bspring%7D%20%3E%20mg)
![Ks > mg](https://tex.z-dn.net/?f=Ks%20%3E%20mg)
Answer:
Average Velocity = 3.65 m/s
Explanation:
Average Velocity
![=\frac{22.1-3.85}{5}=\frac{18.25}{5}=3.65](https://tex.z-dn.net/?f=%3D%5Cfrac%7B22.1-3.85%7D%7B5%7D%3D%5Cfrac%7B18.25%7D%7B5%7D%3D3.65)