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 :
Answer:
%N = 25.94%
%O = 74.06%
Explanation:
Step 1: Calculate the mass of nitrogen in 1 mole of N₂O₅
We will multiply the molar mass of N by the number of N atoms in the formula of N₂O₅.
m(N): 2 × 14.01 g = 28.02 g
Step 2: Calculate the mass of oxygen in 1 mole of N₂O₅
We will multiply the molar mass of O by the number of O atoms in the formula of N₂O₅.
m(O): 5 × 16.00 g = 80.00 g
Step 3: Calculate the mass of 1 mole of N₂O₅
We will sum the masses of N and O.
m(N₂O₅) = m(N) + m(O) = 28.02 g + 80.00 g = 108.02 g
Step 4: Calculate the percent composition of N₂O₅
We will use the following expression.
%Element = m(Element)/m(Compound) × 100%
%N = m(N)/m(N₂O₅) × 100% = 28.02 g/108.02 g × 100% = 25.94%
%O = m(O)/m(N₂O₅) × 100% = 80.00 g/108.02 g × 100% = 74.06%
Answer:
Explanation:
2NO3, Ca2+
Spectator ions exists in both sides of the equation. These ions are present in both sides thus cancel out (they do not contribute to reaction...i.e. they only watch or spectate ;) )
Answer:9.18 m/s
Explanation:
The average speed for the entire trip is found ......total distance/total time. Remember r*t=d, so divide both sides by t and get r = d/t.
So the cyclist went 800+500+1200 m = 2500 m for total distance.
10t =800 leads to t=80 sec
5t=500 leads to t=100 sec
13t = 1200 leads to t = 92.3 sec
total time is 272.3 sec
Average speed for the entire trip is 2500 / 272.3 = 9.18 m/s
Answer:
it is 1,4
Explanation:
Just took the test hope you get it right. c:
