Answer:
The correct answer is 28.2 %.
Explanation:
Based on the given question, the partial pressures of the gases present in the trimix blend is 55 atm oxygen, 50 atm helium, and 90 atm nitrogen. Therefore, the sum of the partial pressure of gases present in the blend is,
Ptotal = PO2 + PN2 + PHe
= 55 + 90 + 50
= 195 atm
The percent volume of each gas in the trimix blend can be determined by using the Amagat's law of additive volume, that is, %Vx = (Px/Ptot) * 100
Here Px is the partial pressure of the gas, Ptot is the total pressure and % is the volume of the gas. Now,
%VO2 = (55/195) * 100 = 28.2%
%VN2 = (90/195) * 100 = 46%
%VHe = (50/195) * 100 = 25.64%
Hence, the percent oxygen by volume present in the blend is 28.2 %.
The atomic mass of a certain element is summation of the product of the decimal equivalent of the percentage abundance and the given atomic mass of each of the isotope. If we let x be the percentage abundance of the 86 amu-isotope then, the second one is 1-x such that,
x(86) + (1 - x)(90) = 87.08
The value of x from the equation is 0.73. This value is already greater than 0.5. Thus, the isotope with greatest abundance is that which is 86 amu.
Answer:
1.18 V
Explanation:
The given cell is:
Half reactions for the given cell follows:
Oxidation half reaction:
Reduction half reaction:
Multiply Oxidation half reaction by 2 and Reduction half reaction by 3
Net reaction:
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
To calculate the
of the reaction, we use the equation:
Putting values in above equation, we get:
To calculate the EMF of the cell, we use the Nernst equation, which is:
where,
= electrode potential of the cell = ?V
= standard electrode potential of the cell = +1.21 V
n = number of electrons exchanged = 6
Putting values in above equation, we get:

Answer:
Since in a chloride ion, we have an additional electron
you might think that it will affect the mass but the mass of an electron is almost negligible so we will ignore that
Amount of ions in 1 mol = 6.022 * 10^23
Amount of ions in 0.486 moles = 0.486 * (6.022*10^23)
Amunt of ions in 0.486 moles = 2.9 * 10^23 ions
Hence, option 1 is correct