Answer:
14175 j heat released.
Explanation:
Given data:
Mass of aluminium = 350.0 g
Initial temperature = 70.0°C
Final temperature = 25.0°C
Specific heat capacity of Aluminium = 0.9 j/g.°C
Heat changed = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
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
Heat change:
ΔT = Final temperature - initial temperature
ΔT = 25.0°C - 70°C
ΔT = -45°C
Q = m.c. ΔT
Q = 350 g × 0.9 j/g.°C × -45°C
Q = -14175 j
The number of particles (molecules or atoms) is: 6.022 x 10²³ particles (atoms or molecules).
1 mol of H₂O has 6.022 x 10²³ molecules.
1 mol of Al has 6.022 x 10²³ atoms.
Help 2 what the answer for this?
Answer: The first isotope has a relative abundance of 79% and last isotope has a relative abundance of 11%
Explanation: Given that the average atomic mass(M) of magnesium
= 24.3050amu
Mass of first isotope (M1) = 23.9850amu
Mass of middle isotope (M2)=24.9858amu
Mass of last isotope(M3)= 25.9826amu
Total abundance = 1
Abundance of middle isotope = 0.10
Let abundance of first and last isotope be x and y respectively.
x+0.10+y =1
x = 0.90-y
M = M1 × % abundance of first isotope + M2 × % of middle isotope +M3 ×% of last isotope
24.03050= 23.985× x + 24.9858 ×0.10 + 25.9826×y
Substitute x= 0.90-y
Then
y = 0.11
Since y=0.11, then
x= 0.90-0.11
x=0.79
Therefore the relative abundance of the first isotope = 11% and the relative abundance of the last isotope = 79%
