As number of gaseous moles in reactant and prodict are same that is 4
So No change will occur
Answer:
127.0665 amu
Explanation:
Firstly, to answer the question correctly, we need to access the percentage compositions of the iodine and the contaminant iodine. We can do this by placing their individual masses over the total and multiplying by 100%.
We do this as follows. Since the mass of the contaminant iodine is 1.00070g, the mass of the 129I in that particular sample will be 12.3849 - 1.00070 = 11.3842g
The percentage abundances is as follows:
Synthetic radioisotope % = 1.0007/12.3849 * 100% = 8.1%
Since there are only two constituents, the percentage abundance of the 129I would be 100 - 8.1 = 91.9%
Now, we can use these percentages to get the apparent atomic mass. We get this by multiplying the percentage abundance’s by the atomic masses of both and adding together.
That is :
[8.1/100 * 128.9050] + [91.9/100 * 126.9045] = 10.441305 + 116.6252355 = 127.0665 amu
Answer:
Molecular formula for the gas is: C₄H₁₀
Explanation:
Let's propose the Ideal Gases Law to determine the moles of gas, that contains 0.087 g
At STP → 1 atm and 273.15K
1 atm . 0.0336 L = n . 0.082 . 273.15 K
n = (1 atm . 0.0336 L) / (0.082 . 273.15 K)
n = 1.500 × 10⁻³ moles
Molar mass of gas = 0.087 g / 1.500 × 10⁻³ moles = 58 g/m
Now we propose rules of three:
If 0.580 g of gas has ____ 0.480 g of C _____ 0.100 g of C
58 g of gas (1mol) would have:
(58 g . 0.480) / 0.580 = 48 g of C
(58 g . 0.100) / 0.580 = 10 g of H
48 g of C / 12 g/mol = 4 mol
10 g of H / 1g/mol = 10 moles
Answer:
ΔG°rxn = -72.9 kJ
Explanation:
Let's consider the following reaction.
HCN(g) + 2 H₂(g) → CH₃NH₂(g)
We can calculate the standard Gibbs free energy of the reaction (ΔG°rxn) using the following expression:
ΔG°rxn = ΔH° - T.ΔS°
where,
ΔH° is the standard enthalpy of the reaction
T is the absolute temperature
ΔS° is the standard entropy of the reaction
ΔG°rxn = -158.0 KJ - 387 K × (-219.9 × 10⁻³ J/K)
ΔG°rxn = -72.9 kJ
