Answer:
Option C. ⁰₊₁e
Explanation:
From the question given above, the following data were obtained:
¹⁵₈O —> ¹⁵₇N + __
Let the unknown be ʸₓM
Thus, the equation becomes
¹⁵₈O —> ¹⁵₇N + ʸₓM
Next, we shall determine x, y and M. This can be obtained as follow:
8 = 7 + x
Collect like terms
8 – 7 = x
1 = x
x = 1
15 = 15 + y
Collect like terms
15 – 15 = y
0 = y
y = 0
ʸₓM => ⁰₁M => ⁰₊₁e
Therefore,
¹⁵₈O —> ¹⁵₇N + ʸₓM
¹⁵₈O —> ¹⁵₇N + ⁰₊₁e
391.12 *
mass is required for activity of 315 ci
The activity of the radioactive substance is the number of disintegrations taking place in a radioactive sample.
Isotopes are the members of an element having the same number of protons but the different number of neutrons.
the activity of the radioactive sample is given as:
A = λ * N
where λ is the decay constant and N is the number of disintegrations.
activity is given as 315 ci
A = 315 * 3.7 * 
= 1165.5 *
Bq
λ = 0.693 / 
= 0.693 / 2.69 * 24 * 3600
= 2.98 *
t^-1
Therefore, N = A / λ
= 1165.5 *
/ 2.98 * 
N = 391.12 * 
391.12 *
mass is required for activity of 315 ci
For more information click on the link below:
brainly.com/question/25746629
#SPJ4
mass of car =700kg
Required acceleration =5m/s^2
According to newton's second law
F=ma
F=(700)(5)
F=3500 Kgm/sec^2
aumenta su velocidad de 60 a 100 Km/h en 20 segundos. Calcular la fuerza resultante que actúa sobre el coche y el espacio recorrido en ese tiempo