Answer is: the average atomic mass 217.606 amu.
Ar₁= 203.973 amu; the average atomic mass of isotope.
Ar₂ = 205.9745 amu.
Ar₃ = 206.9745 amu.
Ar₄ = 207.9766 amu.
ω₁ = 1.40% = 0.014; mass percentage of isotope.
ω₂ = 24.10% = 0.241.
ω₃ = 22.10% = 0.221.
ω₄ = 57.40% = 0.574.
Ar = Ar₁ · ω₁+ Ar₂ · ω₂ + Ar₃ · ω₃ + Ar₄ · ω₄.
Ar = 203.973 amu · 0.014 + 205.9745 amu · 0.241 + 206.9745 amu · 0.221 + 207.9766 amu · 0.574.
Ar = 2.855 amu + 49.632 amu + 45.741 amu + 119.378 amu.
Ar = 217.606 amu.
But abundance of isotopes is greater than 100%.
It should be lead, with the fourth isotope weighs 207.9766 amu and an abundance of 52.40.
Answer:
B) The molecules were closer together when the juice pop was frozen.
Answer:
Explanation:
Hello there!
In this case, according to this calorimetry problem on equilibrium temperature, it is possible for us to infer that the heat released by the metal allow is absorbed by the water for us to write:
Thus, by writing the aforementioned in terms of mass, specific heat and temperature, we have:
Then, we solve for specific heat of the metallic alloy to obtain:
Thereby, we plug in the given data to obtain:
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<h3>
Answer:</h3>
1.93 g
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Explanation:</h3>
<u>We are given;</u>
The chemical equation;
2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(l) ΔH = -3120 kJ
We are required to calculate the mass of ethane that would produce 100 kJ of heat.
- 2 moles of ethane burns to produce 3120 Kilo joules of heat
Number of moles that will produce 100 kJ will be;
= (2 × 100 kJ) ÷ 3120 kJ)
= 0.0641 moles
- But, molar mass of ethane is 30.07 g/mol
Therefore;
Mass of ethane = 0.0641 moles × 30.07 g/mol
= 1.927 g
= 1.93 g
Thus, the mass of ethane that would produce 100 kJ of heat is 1.93 g
