About one kilogram of coffee beans is required
Answer:
V₂ = 0.62 L
Explanation:
Given data:
Initial volume = 2.4 L
Initial temperature = 25°C
Final temperature = -196°C
Final volume = ?
Solution:
Initial temperature = 25°C (25+273 = 298 K)
Final temperature = -196°C ( -196+273 = 77 K)
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 2.4 L × 77 K / 298 k
V₂ = 184.8 L.K / 298 K
V₂ = 0.62 L
Answer:
3.43 %
Explanation:
We need to calculate first the number of moles of CeO2 produced in the combustion. Given its formula we know how many moles of Ce atom are present. From there calculate the mass this number of moles this represent and then one can calculate the percentage.
0.1848 g CeO2 x 1 mol CeO2/172.114g = 0.00107 mol CeO2
0.00107 mol CeO2 x 1 mol Ce/ 1 mol CeO2 = 0.00107 mol Ce
.00107 mol Ce x 140.116 g Ce/ mol = 0.150 g Ce
0.150 g Ce/ 4.3718 g sample x 100 = 3.43 %
The statement above is FALSE.
Unlabeled atom joined to carbon atoms which are not directly part of a ring structure are assumed to be CARBON ATOMS. In a ring structure, an unlabeled atom at the angle where two lines joined together is always assumed to be a carbon atom<span />
