60m27Co → 6027Co Predict the type of radioactive emission produced from the decay of metastable cobalt-60 to cobalt-60. Describe
1 answer:
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The only balanced equation is B.
If you look at the equation and break it down you can see that in:
→ 
Starting from the left side of the equation there are 2 Nitrogen atoms, and 2 oxygen atoms as indicated by the subscript.
To balance the equation, the number of atoms of each element in the right side equation should be equal to left. By putting the numerical coefficient of 2, you will distribute that to each element. So you will end up with 2 nitrogen atoms and 2 oxygen atoms on the left side of the equation. Thus, the equation is balanced.
The answer again, is B.
If you look at the equation and break it down you can see that in:
Starting from the left side of the equation there are 2 Nitrogen atoms, and 2 oxygen atoms as indicated by the subscript.
To balance the equation, the number of atoms of each element in the right side equation should be equal to left. By putting the numerical coefficient of 2, you will distribute that to each element. So you will end up with 2 nitrogen atoms and 2 oxygen atoms on the left side of the equation. Thus, the equation is balanced.
The answer again, is B.
Answer:
mph
Explanation:
= Speed of bird in still air
= Speed of wind = 44 mph
Consider the motion of the bird with the wind
= distance traveled with the wind = 9292 mi
= time taken to travel the distance with wind
Time taken to travel the distance with wind is given as
eq-1
Consider the motion of the bird with the wind
= distance traveled against the wind = 6060 mi
= time taken to travel the distance against wind
Time taken to travel the distance against wind is given as
eq-2
As per the question,
Time taken with the wind = Time taken against the wind
mph
Answer:
t= 27.38 mins [this the time taken by the enzyme to hydrolyse 80% of the fat present]
Explanation:given values
Half life of lipase t_1/2 = 8 min x 60s/min = 480 s
Rate constant for first order reaction
k_d = 0.6932/480 = 1.44 x 10^-3 s-1
Initial fat concentration S_0 = 45 mol/m3 = 45 mmol/L
rate of hydrolysis Vm0 = 0.07 mmol/L/s
Conversion X = 0.80
Final concentration S = S_0(1-X) = 45 (1-0.80) = 9 mol/m3
K_m = 5mmol/L
time take is given by
all values are given and putting these value we get
t=1642.83 secs
which is equal to
t= 27.38 mins [this the time taken by the enzyme to hydrolyse 80% of the fat present]