<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.
84.34 grams of grams of iron (III) chloride that can be produced is maximum because Fe is the limiting reagent in this reaction and chlorine gas is excess reagent.
Explanation:
Balanced chemical equation:
2 Fe + 3 Cl2 → 2 FeCl3
DATA GIVEN:
iron = atoms
mass of chlorine gas = 67.2 liters
mass of FeCl3 = ?
number of moles of iron will be calculated as
number of moles = 
number of moles = 
number of moles = 0.52 moles of iron
moles of chlorine gas
number of moles = 
Putting the values in the equation:
n = 
(atomic mass of chlorine gas = 70.96 grams/mole)
= 947.01 moles
Fe is the limiting reagent so
2 moles of Fe gives 2 moles of FeCl3
0.52 moles of Fe will give
= 
0.52 moles of FeCl3 is formed.
to convert it into grams:
mass = n X atomic mass
= 0.52 x 162.2 (atomic mass of FeCl3 is 162.2grams/mole)
<h3> = 84.34 grams </h3>
The corect answers will be:
1) A
2) B
3) D
Hope this helped :)
[ H₃O⁺] = 10 ^ - pH
[ H₃O⁺ ] = 10 ^ - 7.30
[ H₃O⁺ ] = 5.011 x 10⁻⁸ M
hope this helps!
We first need to find the number of moles of gas in the container
PV = nRT
where;
P - pressure - 2.87 atm x 101 325 Pa/atm = 290 802.75 Pa
V - volume - 5.29 x 10⁻³ m³
n - number of moles
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 230 K
substituting these values in the equation
290 802.75 Pa x 5.29 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 230 K
n = 0.804 mol
the molar mass = mass present / number of moles
molar mass of gas = 56.75 g / 0.804 mol
therefore molar mass is 70.6 g/mol
