The original radioisotope underwent double-alpha decay, where the original nucleus lost a total of 4 protons and 4 neutrons.
The original radioisotope would have an atomic number of 86+4 (protons), and would have an atomic mass of 222+8 (protons + neutrons). This would make the element Thorium-230, or 90Th230.
Answer:
36°C
Explanation:
Given parameters:
Mass of aluminum = 725g
Quantity of heat = 2.35 x 10⁴J
Unknown:
Temperature change = ?
Solution:
To solve this problem, we simply use the expression below:
The quantity of energy is given as:
Q = m C Δt
Q is the quantity of energy
m is the mass
C is the specific heat capacity of aluminum = 0.9J/g°C
Δt is the change in temperature
The unknown is Δt;
Δt =
=
= 36°C
The answer is chemical. Flammability is a chemical property
Answer:
Explanation:
Half life of Nobelium-253 is 97 seconds . That means after every 97 seconds half of the Nobelium amount will be disintegrated .
Time taken in bringing the sample to laboratory = 291 seconds
291 second = 291 / 97 half life
n = 3
N = 
N₀ is original mass , N is mass after n number of half life.
N = 5 mg x 
= .625 mg
Only 0.625 mg of Nobelium-253 will be left .
NaOH + HNO₃ --------> NaNO₃ + H₂O
___________________
NaNO₃ - sodium nitrate