Thermodynamics, Nuclear Physics, Quantum Physics, Astronomy and Astrophysics
Answer:
P = 0.0009417 atm
Or,
P = 9.417 × 10⁻⁴ atm
Or,
P = 0.0954157 kPa
Or,
P = 0.715677 mmHg (Torr)
Explanation:
Data Given:
Moles = n = 3.2 mol
Temperature = T = 312 K
Pressure = P = ?
Volume = V = 87 m³ = 87000 L
Formula Used:
Let's assume that the gas is acting as an Ideal gas, the according to Ideal Gas Equation,
P V = n R T
where; R = Universal Gas Constant = 0.082057 atm.L.mol⁻¹.K⁻¹
Solving Equation for P,
P = n R T / V
Putting Values,
P = (3.2 mol × 0.082057 atm.L.mol⁻¹.K⁻¹ × 312 K) ÷ 87000 L
P = 0.0009417 atm
Or,
P = 9.417 × 10⁻⁴ atm
Or,
P = 0.0954157 kPa
Or,
P = 0.715677 mmHg (Torr)
1Hz = 1 cycle per second
19 cycles / .5 seconds = 38Hz
<u>Answer:</u> The atomic weight of the second isotope is 64.81 amu.
<u>Explanation:</u>
Average atomic mass of an element is defined as the sum of atomic masses of each isotope each multiplied by their natural fractional abundance
Formula used to calculate average atomic mass follows:
.....(1)
We are given:
Let the mass of isotope 2 be 'x'
Mass of isotope 1 = 62.9 amu
Percentage abundance of isotope 1 = 69.1 %
Fractional abundance of isotope 1 = 0.691
Mass of isotope 2 = 'x'
Percentage abundance of isotope 2 = 30.9%
Fractional abundance of isotope 2 = 0.309
Average atomic mass of copper = 63.5 amu
Putting values in equation 1, we get:
![\text{Average atomic mass of copper}=[(62.9\times 0.691)+(x\times 0.309)]](https://tex.z-dn.net/?f=%5Ctext%7BAverage%20atomic%20mass%20of%20copper%7D%3D%5B%2862.9%5Ctimes%200.691%29%2B%28x%5Ctimes%200.309%29%5D)

Hence, the atomic weight of second isotope will be 64.81 amu.