<h3>
Answer:</h3>
B. 0.33 mol
<h3>
Explanation:</h3>
We are given;
Gauge pressure, P = 61 kPa (but 1 atm = 101.325 kPa)
= 0.602 atm
Volume, V = 5.2 liters
Temperature, T = 32°C, but K = °C + 273.15
thus, T = 305.15 K
We are required to determine the number of moles of air.
We are going to use the concept of ideal gas equation.
- According to the ideal gas equation, PV = nRT, where P is the pressure, V is the volume, R is the ideal gas constant, (0.082057 L.atm mol.K, n is the number of moles and T is the absolute temperature.
- Therefore, to find the number of moles we replace the variables in the equation.
- Note that the total ball pressure will be given by the sum of atmospheric pressure and the gauge
- Therefore;
- Total pressure = Atmospheric pressure + Gauge pressure
We know atmospheric pressure is 101.325 kPa or 1 atm
Total ball pressure = 1 atm + 0.602 atm
= 1.602 atm
That is;
PV = nRT
n = PV ÷ RT
therefore;
n = (1.602 atm× 5.2 L) ÷ (0.082057 × 305.15 K)
= 0.3326 moles
= 0.33 moles
Therefore, there are 0.33 moles of air in the ball.
The ionization energy of an element is the amount of energy required to remove one mole of electrons from the element in its gaseous state. The equations for the first three are:
Fe(g) → Fe⁺(g) + e⁻
Fe⁺(g) → Fe⁺²(g) + e⁻
Fe⁺²(g) → Fe⁺³(g) + e
Answer:
A = 674.33mmHg
B = 0.385atm
Explanation:
Both question A and B requires the application of pressure law which states that the pressure of a fixed mass of gas is directly proportional to its temperature provided that volume is kept constant.
Mathematically,
P = kT, k = P / T
P1 / T1 = P2 / T2 = P3 / T3 =.......= Pn/Tn
A)
Data:
P1 = 799mmHg
T1 = 50°C = (50 + 273.15) = 323.15K
P2 = ?
T2 = 273.15K
P1 / T1 = P2 / T2
Solve for P2
P2 = (P1 × T2) / T1
P2 = (799 × 273.15) / 323.15
P2 = 674.37mmHg
The final pressure is 674.37mmHg
B)
P1 = 0.470atm
T1 = 60°C = (60 + 273.15)K = 333.15K
P2 = ?
T2 = 273.15K
P1 / T1 = P2 / T2
Solve for P2,
P2 = (P1 × T2) / T1
P2 = (0.470 × 273.15) / 333.15
P2 = 0.385atm
The final pressure is 0.385atm
Answer: 
Explanation:
Significant figures : The figures in a number which express the value or the magnitude of a quantity to a specific degree of accuracy is known as significant digits.
Rules for significant figures:
Digits from 1 to 9 are always significant and have infinite number of significant figures.
All non-zero numbers are always significant.
All zero’s between integers are always significant.
All zero’s after the decimal point are always significant.
All zero’s preceding the first integers are never significant.
Thus has three significant figures
