Answer:
I have no clue what the question is
Answer:
ⁿₐX => ²¹⁸₈₄Po
Explanation:
Let ⁿₐX be the isotope.
Thus, the equation can be written as follow:
²²²₈₆Rn —> ⁴₂α + ⁿₐX
Next, we shall determine the value of 'n' and 'a'. This can be obtained as follow:
222 = 4 + n
Collect like terms
222 – 4 = n
218 = n
Thus,
n = 218
86 = 2 + a
Collect like terms
86 – 2 = a
84 = a
Thus,
a = 84
ⁿₐX => ²¹⁸₈₄Po
²²²₈₆Rn —> ⁴₂α + ⁿₐX
²²²₈₆Rn —> ⁴₂α + ²¹⁸₈₄Po
Answer:
Q = 10.8 KJ
Explanation:
Given data:
Mass of Al= 100g
Initial temperature = 30°C
Final temperature = 150°C
Heat required = ?
Solution:
Specific heat of Al = 0.90 j/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 150°C - 30°C
ΔT = 120°C
Q = 100g×0.90 J/g.°C× 120°C
Q = 10800 J (10800j×1KJ/1000 j)
Q = 10.8 KJ
At the melting point. Draw a line up from 0 degrees and a line to the right from 1 atm. They meet at the line between solid and liquid... the melting point