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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If you mean the SI Unit of GPE, the answer is J for Joules.
if that's not the question being asked, i would need a little more elaboration please :)
Answer:
E = 3.54 x 10⁻¹⁹ J
Explanation:
The energy of the photon can be given in terms of its wavelength by the use of the following formula:

where,
E = energy = ?
h = Plank's Constant = 6.626 x 10⁻³⁴ Js
c = speed of light = 2.998 x 10⁸ m/s
λ = wavelength of light = 560.6 nm = 5.606 x 10⁻⁷ m
Therefore,

<u>E = 3.54 x 10⁻¹⁹ J</u>
There are some missing data in the text of the problem. I've found them online:
a) coefficient of friction dry steel piston - steel cilinder: 0.3
b) coefficient of friction with oil in between the surfaces: 0.03
Solution:
a) The force F applied by the person (300 N) must be at least equal to the frictional force, given by:

where

is the coefficient of friction, while N is the normal force. So we have:

since we know that F=300 N and

, we can find N, the magnitude of the normal force:

b) The problem is identical to that of the first part; however, this time the coefficienct of friction is

due to the presence of the oil. Therefore, we have:
The sun's energy is refferd to solor energy