Electron affinity for fluorine is than chlorine most likely , due to the electron repulsion that occur between the electron where n= 2 . the elements in the second period have such small electron clouds that electron repulsion is greater than that of the rest of the family.
Answer:
The 12L helium tank pressurized to 160 atm will fill <em>636 </em>3-liter balloons
Explanation:
It is possible to answer this question using Boyle's law:
Where P₁ is the pressure of the tank (160atm), V₁ is the volume of the tank (12L), P₂ is the pressure of the balloons (1atm, atmospheric pressure) And V₂ is the volume this gas will occupy at 1 atm, thus:
160atm×12L = 1atm×V₂
V₂ = 1920L
As the tank will never be empty, the volume of the gas able to fill balloons is the total volume minus 12L, thus the volume of helium able to fill balloons is:
1920L - 12L = 1908L
1908L will fill:
1908L×
= <em>636 balloons</em>
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I hope it helps!
Answer:
The answer to your question is: 68.3%
Explanation:
Urea = ?
Theoretical yield
Ammonia 145.1 kg
Carbon dioxide = 211.4 kg
Urea = 174.9 kg
MW NH3 = 17kg
MW CO2 44 kg
MW Urea = 60 kg
2NH₃(aq) + CO₂(aq) → CH₄N₂O(aq) + H₂O(l)
34 44 kg 60kg
145.1 211.4 174.9
Limiting reactant = NH3
34kg of NH3 ------------------ 60 kg of Urea
145.1 kg ------------------- x
x = (145.1 x 60) / 34
x = 256 kg of Urea
Percent yield = 174.9 / 256 x 100
= 68.3%
Answer:
The answer to your question is below
Explanation:
Data
mass of Gold = 35.12 g
mass number of gold = 197 g (from the periodic table)
1.- To calculate the number of moles use proportions
197 g of gold ----------- 1 mol
35.12 g of gold -------- x
x = (35.12 x 1)/197
<u>x = 0.18 moles of gold</u>
2.- To calculate the number of atoms also use proportions and the Avogadro's number.
1 mol of gold ------------ 6.023 x 10 ²³ atoms
0.18 moles of gold ----- x
x = (0.18 x 6.023 x 10²³) / 1
<u> x = 1.084 x 10²³ atoms</u>
