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2 Na₂CO₃ + 2 CuSO₄ + H₂O → CuCO₃.Cu(OH)₂ + 2 Na₂SO₄ + CO₂
Malachite molar mass = 221.1 g / mol
So 2 moles CuSO₄ produce 1 mole of malachite
so 1.5 mole CuSO₄ produce (0.75) mole malachite
Mass of malachite = 0.75 mole * 221.1 g/ mol = 165.83 g
<span>R= 8.314 J/mol K
T= 273 + 102 = 375K</span><span>
= (3/2) x 8.314 x 375 = 4680 J/mol</span>
Bone age : 22,920 years
<h3>Further explanation</h3>
Given
Nt = 2.5 g C-14
No = 40 g
half-life = 5730 years
Required
time of decay
Solution
General formulas used in decay:

t = duration of decay
t 1/2 = half-life
N₀ = the number of initial radioactive atoms
Nt = the number of radioactive atoms left after decaying during T time
Input the value :

To find AH°rxn, we use the following equation:
What we're going to do is to sum the enthalpy of the products and then substract with the enthalpy of the reactives:
As you can see, we need to multiply by the coefficients of the reaction.
Now, just replace the values of the table:
So the answer is -822.2kJ/mol.
For b:
Now, just replace the values of the table:
The answer for b is -1036kJ/mol.