163 lb * 1 kg / 2.205 lb * 15.0 mg/kg = 1108.8 mg or about 1.11 g
Delta H of solution = -Lattice Energy + Hydration
<span>Delta H of solution=- (-730)+(-793) </span>
<span>Delta H of solution= -63kJ/mol </span>
<span>Now we find moles of LiI: </span>
<span>10gLiI/133.85g=.075moles </span>
<span>multiply moles to the delta H of solution to cross cancel moles. .75moles x -64kJ/mol =4.7</span>
Bicarbonate, sodium, potassium
are regulated by the kidneys.
Answer:
The percent isotopic abundance of Ir-193 is 60.85 %
The percent isotopic abundance of Ir-191 is 39.15 %
Explanation:
we know there are two naturally occurring isotopes of iridium, Ir-191 and Ir-193
First of all we will set the fraction for both isotopes
X for the isotopes having mass 193
1-x for isotopes having mass 191
The average atomic mass is 192.217
we will use the following equation,
193x + 191(1-x) = 192.217
193x + 191 - 191x = 192.217
193x- 191x = 192.217 - 191
2x = 1.217
x= 1.217/2
x= 0.6085
0.6085 × 100 = 60.85 %
60.85% is abundance of Ir-193 because we solve the fraction x.
now we will calculate the abundance of Ir-191.
(1-x)
1-0.6085 =0.3915
0.3915× 100= 39.15 %
