Answer:
Percentage yield = 85.2%
Explanation:
Given data:
Mass of Mg = 21.3 g
Actual yield of MgO = 30.2 g
Percentage yield = ?
Solution:
Chemical equation:
2Mg + O₂ → 2MgO
Number of moles of Mg = mass/molar mass
Number of moles of Mg = 21.3 g / 24.3 g/mol
Number of moles of Mg = 0.88 mol
Now we will compare the moles of MgO with Mg.
Mg : MgO
2 : 2
0.88 : 0.88
Mass of MgO:
Mass of MgO= moles × molar mass
Mass of MgO= 0.88 mol × 40.3g/mol
Mass of MgO = 35.46 g
Actual yield of MgO = 30.2 g
Percentage yield:
Percentage yield = Actual yield/theoretical yield × 100
Percentage yield = 30.2 g/ 35.46 g × 100
Percentage yield = 85.2%
Answer:
sulfur promotes oxide-reduction reactions.
Explanation:
In stagnant water, some solutes tend to precipitate. When Sulfur precipitate and touch a metal, Sulfur is being reduced and the metal is oxidated. This depends of potential redox of each element.
Answer:
14 moles
Explanation:
For an Ideal gas,
PV = nRT...................... Equation 1
Where P = Pressure, V = Volume, n = number of mole, R = Molar gas constant.
make n the subject of the equation
n = PV/RT.................. Equation 2
Given: P = 14.297 atm, V = 22.9 L = 22.9 dm³, T = 12 °C = (12+273) K = 285 K.
Constant: R = 0.082 atm.dm³/K.mol
Substitute these values into equation 2
n = (14.297×22.9)/(285×0.082)
n = 327.4013/23.37
n = 14.009
n ≈ 14 moles
The answer is A.4m/s because a bowling ball has more mass then a basketball
I hope this helps
