<span>4 x 36 moles. of Phosporus and 10 x 36 of Oxygen. I hope this helps. (:</span>
Mass defect for oxygen-16 = 0. 13261 amu, in the kilograms the mass defect equals to 2.20 × 10⁻²⁸ kg.
<h3>What is mass defect?</h3>
Mass defect is the difference between the mass of of an whole atom and the combined mass of its individual particles present in that atom.
We know that, 1 amu = 1.6 × 10⁻²⁷ kg
Given that, mass defect for oxygen-16 = 0.13261 amu
To calculate this defect in terms of kilograms, we have to convert into kg unit as:
0.13261 amu = 0.13261 amu × 1.6 × 10⁻²⁷ kg/amu
0.13261 amu = 2.20 × 10⁻²⁸ kg
Hence option (2) is correct.
To know more about Mass defect, visit the below link:
brainly.com/question/4334375
<span>because p6 will be the group 8. You have to count the 2 electrons from the "s" block that are Group I and Group II
Group I s1
Group II s2
Group III s2 p1
Group IV s2 p2
Group V s2 p3
Group VI s2 p4
Group VII s2 p5
Group VIII s2 p6</span>
Answer:
The final pressure in the container at 0°C is 2.49 atm
Explanation:
We apply the Ideal Gases law to know the global pressure.
We need to know, the moles of each:
P He . V He = moles of He . R . 273K
(1atm . 4L) / R . 273K = moles of He → 0.178 moles
P N₂ . V N₂ = moles of N₂ . R . 273K
(1atm . 6L) / R . 273K = moles of N₂ → 0.268 moles
P Ar . V Ar = moles of Ar . R . 273K
(1atm . 10L) / R . 273K = moles of Ar → 0.446 moles
Total moles: 0.892 moles
P . 8L = 0.892 mol . R . 273K
P = ( 0.892 . R . 273K) / 8L = 2.49 atm
R = 0.082 L.atm/mol.K
