The electron geometry is tetrahedral and the molecular geometry is tetrahedral. If a molecule of CH3OCH3 is to be drawn, the two carbons would have four single bonds and the middle O would have two single bonds and the two lone pairs. Molecular geometry does not consider the lone pairs as bonds like in electron domain geometry. However, since the carbons do not contain any lone pairs, its electrons domain and molecular geometry will be the same. Therefore; Both carbons are tetrahedral for electron domain geometry and molecular geometries. The O is tetrahedral for the electron domain geometry and bent for molecular geometry.
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.
The correct answer is B. Adipic Acid.
According to coursehero.com
Ca(OH)2
Make sure your charges are balanced when you write the compound. Ca has a +2 charge and OH has a -1 charge so you need 2 of OH.
Answer:
3.5 mol·L⁻¹
Explanation:
1. Set up an ICE table.
2. Solve for x
The equilibrium concentration of NO is 1.0 mol·L⁻¹, so
1.0 = 2.0 - 2x
2x + 1.0 = 2.0
2x = 1.0
x = 0.5
3. Calculate the equilibrium concentration of I₂
[I₂] = 4.0 - x = 4.0 - 0.5 = 3.5 mol·L⁻¹
