A 13.58 g sample of a compound contains 8.67 g of iron, Fe, 1.60 g of phosphorus, P, and oxygen, O. Calculate the empirical
1 answer:
Answer:
Explanation:
To do this, we find the moles of each element. We get around 0.155 moles of Fe, 0.051 moles of P, and 0.206 moles of O. We then divide each one by the smallest one (which is 0.051 moles of P). We then get 3 for Fe, 1 for P, and 4 for O. This correlates to the empirical formula of the compound.
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Answer:
Final volumeof the gas = 2.84 L
Explanation:
The formular to be used here is the general gas equation. the formular is being used because it gives the relationship between the three gas parameters (volume, temperature and pressure) mentioned.
The general gas equation is given as;
where;
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = Final pressure
V2 = Final volume
T2 = Final temperature
From the question,
P1 = 1.00 atm
P2 = 0.85atm
T1 = 25C + 273 = 298K (Converting to kelvin)
T2 = 15C + 273 = 288K (Converting to kelvin)
V1 = 2.5L
V2 = ?
from the equation, making V2 subject of formula we have;
V2 = (1*2.5*288)/(298*0.85) = 2.84 L.
Answer: The half-life of a first-order reaction is,
Explanation:
All the radioactive reactions follows first order kinetics.
Rate law expression for first order kinetics is given by the equation:
where,
k = rate constant = ?
t = time taken = 440 s
= initial amount of the reactant = 0.50 M
[A] = left amount = 0.20 M
Putting values in above equation, we get:
The equation used to calculate half life for first order kinetics:
Putting values in this equation, we get:
Therefore, the half-life of a first-order reaction is,