Answer:
The mass percent of potassium is 39%
Option C is correct
Explanation:
Step 1: Data given
Atomic mass of K = 39.10 g/mol
Atomic mass of H = 1.01 g/mol
Atomic mass of C = 12.01 g/mol
Atomic mass of O = 16.0 g/mol
Step 2: Calculate molar mass of KHCO3
Molar mass KHCO3 = 39.10 + 12.01 + 1.01 + 3*16.0
Molar mass KHCO3 = 100.12 g/mol
Step 3: Calculate mass percent of potassium (K)
%K = (atomic mass of K / molar mass of KHCO3) * 100%
%K = (39.10 / 100.12) * 100%
%K = 39.05 %
The mass percent of potassium is 39%
Option C is correct
Answer:
Energy in the campfire originates from the potential chemical energy of the wood, before it is burnt to warm and give light around the campfire.
Explanation:
For a camp fire, the energy input is in the form of the potential chemical energy, stored up in the firewood used to fuel the flame.
The energy output is in the form of heat energy that the campfire radiates all around, light energy given off from the flame, and a little bit of sound energy, heard in the cracking of the firewood as they burn in the flame.
chemical energy ⇒ heat energy + light energy + sound energy
Yes it need to be like that cause when it like that it like that
Answer:
ΔG° = -5.4 kJ/mol
ΔG = 873.2 J/mol = 0.873 kJ /mol
Explanation:
Step 1: Data given
ΔG (NO2) = 51.84 kJ/mol
ΔG (N2O4) = 98.28 kJ/mol
Step 2:
ΔG = ΔG° + RT ln Q
⇒with Q = the reaction quatient
⇒with T = the temperature = 298 K
⇒with R = 8.314 J / mol*K
⇒with ΔG° = ΔG° (N2O4) - 2*ΔG°(NO2
)
⇒ ΔG° = 98.28 kJ/mol - 2* 51.84 kJ/mol
⇒ ΔG° = -5.4 kJ/mol
Part B
ΔG = ΔG° =RT ln Q
⇒with G° = -5.4 kj/mol = -5400 j/mol
⇒
with R = 8.314 J/K*mol
⇒with T = 298 K
⇒with Q = p(N2O4)/ [ p(NO2) ]² = 1.63/0.36² = 12.577
ΔG = -5400 + 8.314 * 298 * ln(12.577)
ΔG = -5400 + 8.314 * 298 * 2.532
ΔG = 873.2 J/mol = 0.873 kJ/mol
