Answer:
88%.
Explanation:
The percentage yield of lead sulfate in this experiment is 88% if 2.53 is divided by 2.85 and multiply by 100. The percentage yield can be calculated when the experimental yield is divided by theoretical yield and then multiply by 100. The percentage yield tells us about the actual yield that is gained in the end of experiment which is lower than theoretical yield.
Answer:
204.73K
Explanation:
the formula : PV=nRT
n=4
P=5.6 atm
V=12 L
R=0.08206 L atm mol-1 K-1
T=?
So, if you plug it in, you will get:-
T=PV/nR
T=(5.6 atm)(12 L)/(4 mol)(0.08206 L atm mol-1 K-1)
T=204.73 K
hope this is correct!
If they are in the same row they have the same number of outer electrons on the same main shell also known as valence electrons
if they are in the same column they have the same number of valence electrons but on different main shells
This question can be simply solved by using heat formula,
Q = mCΔT
Q = heat energy (J)
m = Mass (kg)
C = Specific heat capacity (J / kg K)
ΔT = Temperature change (K)
when water freezes, it produces ice at 0°C (273 K)
hence the temperature change is 25 K (298 K - 273 K)
C for water is 4186 J / kg K or 4.186 J / g K
By applying the equation,
Q = 456 g x 4.186 J / g K x 25 K
= 47720.4 J
= 47.72 kJ
hence 47.72 kJ of heat energy should be removed.
The reaction is properly written as
Mg₃N₂ (s) + 3 H₂O (l) --> 2 NH₃<span> (g) + 3 MgO (s)
Molar mass of Mg</span>₃N₂ = 100.95 g/mol
Molar mass of H₂O = 18 g/mol
Molar mass of MgO = 40.3 g/mol
Moles Mg₃N₂: 3.82/100.95 = 0.0378
Moles H₂O: 7.73/18 = 0.429
Theo H₂O required for available Mg₃N₂: 0.0378*3/1 = 0.1134 mol
Hence, the limiting reactant is Mg₃N₂.
Thus,
Theoretical Yield = 0.0378 mol Mg₃N₂ * 3 mol MgO/Mg₃N₂ * 40.3 g/mol
Theo Yield = 4.57 g
Percent Yield = Actual Yield/Theo Yield * 100
Percent Yield = 3.60 g/4.57 g * 100 =<em> 78.77%</em>
