Answer:
Pro exercise con suffication
Explanation:
...
The anion<span> is also </span>larger than<span> the </span>atom<span> because of </span>electron-electron repulsion<span>. As more </span>electrons are<span> added to the </span>outer shell<span>, and even to </span>higher<span> principle energy levels, the </span>repulsion<span> bewteen the negatively charged particles grows, pushing the </span>shells<span> farther from the nucleus.</span>
Answer:
It's 23.14 percent
Explanation:
First, the mass of all the elements are:
N = 14
O = 16
Fe = 56
In this molecule you have 3 atoms of N, and 9 atoms of O, so:
3•14 = 42
16•9 = 144
The whole mass of the molecule is:
56 + 42 + 144 = 242
242/100 = 2.42, so 1% is 2.42
56/2.42 = 23.14%
Answer:
V₂ = 4.7 L
Explanation:
Given data:
Initial volume = 5.0 L
Initial pressure = 1.50 atm
Final pressure = 1240 mmHg (1240/760 = 1.6 atm)
Final volume = ?
Solution:
P₁V₁ = P₂V₂
V₂ = P₁V₁ / P₂
V₂ = 1.50 atm ×5.0 L/1.6 atm
V₂ = 7.5 atm. L /1.6 atm
V₂ = 4.7 L
Silver chloride produced : = 46.149 g
Limiting reagent : CuCl2
Excess remains := 3.74 g
<h3>Further explanation</h3>
Reaction
silver nitrate + copper(II) chloride ⇒ silver chloride + copper(II) nitrate
Required
silver chloride produced
limiting reagent
excess remains
Solution
Balanced equation
2AgNO3 (aq) + CuCl2 (s) → 2AgCl(s) + Cu(NO3)2(aq)
mol AgNO3 :
= 58.5 : 169,87 g/mol
= 0.344
mol CuCl2 :
=21.7 : 134,45 g/mol
= 0.161
mol ratio : coefficient of AgNO3 : CuCl2 :
= 0.344/2 : 0.161/1
= 0.172 : 0.161
CuCl2 as a limiting reagent
mol AgCl :
= 2/1 x 0.161
= 0.322
Mass AgCl :
= 0.322 x 143,32 g/mol
= 46.149 g
mol remains(unreacted) for AgNO3 :
= 0.344-(2/1 x 0.161)
= 0.022
mass AgNO3 remains :
= 0.022 x 169,87 g/mol
= 3.74 g
