464 g radioisotope was present when the sample was put in storage
<h3>Further explanation</h3>
Given
Sample waste of Co-60 = 14.5 g
26.5 years in storage
Required
Initial sample
Solution
General formulas used in decay:
t = duration of decay
t 1/2 = half-life
N₀ = the number of initial radioactive atoms
Nt = the number of radioactive atoms left after decaying during T time
Half-life of Co-60 = 5.3 years
Input the value :
<span>2.40 - 1.68 =0.72 g of oxigen
moles = 0.72/16 g/mol=0.045
moles x = 1.68/ 55.9=0.03
0.03/0.03 = 1 = x
0.045 / 0.03 = 1.5 = O
to get whole numbers multiply by 2
x2O3
X2O3 +3 CO = 2 X + 3 CO2</span>
Answer: the correct answer is C velocity.
Explanation: I just got the answer wrong on the exam.
Answer:
385.69 g of O₂
Solution:
The Balance Chemical equation for said reaction is as follow;
2 H₂ + O₂ → 2 H₂O
According to Equation,
4.032 g ( 2 mol) H₂ reacts to produce = 36.03 g (2 mol) of H₂O
So,
48.6 g H₂ on reaction will produce = X g of H₂O
Solving for X,
X = (48.6 g × 36.03 g) ÷ 4.032 g
X = 434.29 g of H₂O
It means that the H₂ provided is in Excess. Therefore, the yield of product (H₂O) is being controlled by O₂ (Limiting Reagent).
So, According to Equation,
36.03 g (2 mol) H₂O is produced by = 31.998 g (1 mol) of O₂
So,
434.29 g of H₂O will be produced by = X g of O₂
Solving for X,
X = (434.29 g × 31.998 g) ÷ 36.03 g
X = 385.69 g of O₂
