Answer:
0.625 g
Explanation:
Given data:
Half life of radon-222 = 3.8 days
Total mass of sample = 10 g
Mass left after 15.2 days = ?
Solution:
Number of half lives = T elapsed / Half life
Number of half lives = 15.2 / 3.8
Number of half lives = 4
At time zero = 10 g
At first half life = 10 g/2 = 5 g
At 2nd half life = 5 g/ 2= 2.5 g
At third half life = 2.5 g/2 = 1.25 g
At 4th half life = 1.25 g/2 = 0.625 g
Answer and Explanation:
For the following balanced reaction:
PCl₅(g) ↔ PCl₃(g) + Cl₂(g)
We can see that all reactants and products are gases, so it is an homogeneous equilibrium. The expression for the equilibrium constant Kp can be written from the partial pressures (P) of reactants and products as follows:
Where PPCl₃ is the partial pressure of PCl₃ (reactant), PCl₂ is the partial pressure of Cl₂ (reactant) and PPCl₅ is the partial pressure of PCl₅ (product).
The conclusion that would support the students prediction is B) Plant "A" grows taller than Plant "B".
If the student thinks that adding fertilizer to the plant would help it grow then answer B) would make the most sence.
Answer:
10.6 g CO₂
Explanation:
You have not been given a limiting reagent. Therefore, to find the maximum amount of CO₂, you need to convert the masses of both reactants to CO₂. The smaller amount of CO₂ produced will be the accurate amount. This is because that amount is all the corresponding reactant can produce before it runs out.
To find the mass of CO₂, you need to (1) convert grams C₂H₂/O₂ to moles (via molar mass), then (2) convert moles C₂H₂/O₂ to moles CO₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles CO₂ to grams (via molar mass). *I had to guess the chemical reaction because the reaction coefficients are necessary in calculating the mass of CO₂.*
C₂H₂ + O₂ ----> 2 CO₂ + H₂
9.31 g C₂H₂ 1 mole 2 moles CO₂ 44.0095 g
------------------ x ------------------- x ---------------------- x ------------------- =
26.0373 g 1 mole C₂H₂ 1 mole
= 31.5 g CO₂
3.8 g O₂ 1 mole 2 moles CO₂ 44.0095 g
------------- x -------------------- x ---------------------- x -------------------- =
31.9988 g 1 mole O₂ 1 mole
= 10.6 g CO₂
10.6 g CO₂ is the maximum amount of CO₂ that can be produced. In other words, the entire 3.8 g O₂ will be used up in the reaction before all of the 9.31 g C₂H₂ will be used.
