Answer:
The Avogadro's number is 
Explanation:
From the question we are told that
The edge length is 
The density of the metal is 
The molar mass of Ba is 
Generally the volume of a unit cell is

substituting value
![V = [5.02 *10^{-10}]^3](https://tex.z-dn.net/?f=V%20%3D%20%20%5B5.02%20%2A10%5E%7B-10%7D%5D%5E3)
From the question we are told that 68% of the unit cell is occupied by Ba atoms and that the structure is a metal which implies that the crystalline structure will be (BCC),
The volume of barium atom is

substituting value


The Molar mass of barium is mathematically represented as

Where
is the Avogadro's number
So

substituting value


Answer:
it has 0g of sugar (hopes this helps)
There are several information's already given in the question. Based on those information's, the answer can be easily deduced.
Amount of gasoline required by Harry's car to travel 25 miles = 1 gallon
Then
amount of gasoline required
by Harry's car to travel 15000 miles = 15000/25
= 600 gallons
So
Amount of CO2 released by burning 1 gallon of gasoline = 20 pounds
Then
Amount of CO2 released
by burning 600 gallon of gasoline = 600 * 20
= 12000 pounds
From the above deduction, it can be concluded that the amount of CO2 that will be added by Harry's car to the atmosphere is 12000 pounds.
Answer:
See explanation and image attached
Explanation:
A bond line structure refers to any structure of a covalent molecule wherein the covalent bonds present in the molecule are represented with a single line for each level of bond order.
The bond-line structure of CH3CH2O(CH2)2CH(CH3)2 has been shown in the image attached. We know that oxygen has a lone pair of electrons and this has been clearly shown also in the image attached.