mass of PbI₂ = 27.6606 g
<h3>Further explanation</h3>
Given
Pb(NO₃)₂ + NaI → PbI₂ + NaNO₃
28.0 grams of Pb(NO₃)₂ react with 18.0 grams of NaI
Required
mass of PbI₂
Solution
Balanced equation
Pb(NO₃)₂ + 2NaI → PbI₂ + 2NaNO₃
The principle of a balanced reaction is the number of atoms in the reactants = the number of atoms in the product
mol Pb(NO₃)₂ :
= 28 : 331,2 g/mol
= 0.0845
mol NaI :
= 18 : 149,89 g/mol
= 0.12
Limiting reactant : mol : coefficient
Pb(NO₃)₂ : 0.0845 : 1 = 0.0845
NaI : 0.12 : 2 = 0.06
NaI limiting reactant (smaller ratio)
mol PbI₂ based on NaI
= 1/2 x 0.12 = 0.06
Mass PbI₂ :
= 0.06 x 461,01 g/mol
= 27.6606 g
<span>valence electrons is our answer cuz i am good at chemstry</span>
Answer:
2 atoms of H in the products side
Explanation:
3NO₂ + H₂O → NO + 2HNO₃
In product side: 3 N (1 from NO and 6 from nitric acid), 7 O (6 from nitric acid + 1 from NO), 2 H
In reactant side: 3 N, 7 O (6 from the dioxide + 1 from water) and 2 H
In an average mass, each entry has equal weight. In a weighted average, we multiply each entry by a number representing its relative importance.
Assume that your class consists of 15 girls and 5 boys. Each girl has a mass of 54 kg, and each boy has a mass of 62 kg.
<em>Average mass</em> = (girl + boy)/2 = (54 kg + 62 kg)/2 = <em>58 kg</em>
<em>Weighted average (Method 1)
</em>
Use the <em>numbers of each</em> gender (15 girls + 5 boys)
,
Weighted average = (15×54 kg + 5×62 kg)/20 = (810 kg + 310 kg)/20
= 1120 kg/20 = <em>56 kg</em>.
If you put all the students on one giant balance, their total mass would be
1120 kg and the average mass of a student would be <em>56 kg.
</em>
<em>Weighted average (Method 2)
</em>
Use the <em>relative percentages</em> of each gender (75 % girls and 25 % boys).
Weighted average = 0.75×54 kg + 0.25×62 kg = 40.5 kg + 15.5 kg = <em>56 kg</em>
Each girl contributes 40.5 kg and each boy contributes 15.5 kg to the <em>weighted average</em> mass of a student.