A typical sugar cube has an edge length of 1.0 cm. If you had a cubical box that contained 1.7 mol of sugar cubes, what would its edge length be? (One mole = 6.02 x 1023 units.)
and
A severe thunderstorm dumped 2.6 in of rain in 30 min on a town of area 32 km2. What mass of water (in kg) fell on the town? One cubic meter of water has a mass of 103 kg.
To find the edge length of the cubical box, we need to determine the number of sugar cubes it contains first.
Given:
1 mole = 6.02 x 10^23 sugar cubes
1.7 moles of sugar cubes
To find the number of sugar cubes:
Number of sugar cubes = 1.7 moles x (6.02 x 10^23 sugar cubes/mole)
Number of sugar cubes = 1.7 x (6.02 x 10^23)
Number of sugar cubes = 1.7 x 6.02 x 10^23
Number of sugar cubes = 1.0 x 10^24 sugar cubes
Since a cubical box contains 1.0 x 10^24 sugar cubes (given above), each side of the box must contain the same number of sugar cubes, so the total number of sugar cubes is equal to the number of cubes per side cubed:
Number of sugar cubes = (Edge length)^3
Replacing the number of sugar cubes with 1.0 x 10^24, we can solve for the edge length:
1.0 x 10^24 = (Edge length)^3
Taking the cube root of both sides:
Edge length = ∛(1.0 x 10^24)
Edge length ≈ 1.0 x 10^8 cm
Therefore, the edge length of the cubical box that contains 1.7 mol of sugar cubes is approximately 1.0 x 10^8 cm.
Next, to find the mass of water that fell on the town:
Given:
Rainfall = 2.6 in = 2.6 x 2.54 cm = 6.604 cm (since 1 in = 2.54 cm)
Area of town = 32 km^2 = 32 x 10^6 m^2 (since 1 km^2 = 10^6 m^2)
Density of water = 10^3 kg/m^3 (given)
To find the mass of water:
Volume of water = Rainfall x Area of town
Volume of water = 6.604 cm x 32 x 10^6 m^2 = 211,328,000 cm^3 (since 1 m^3 = 1 x 10^6 cm^3)
Converting cm^3 to m^3:
Volume of water = 211,328,000 cm^3 = 211,328 m^3 (since 1 m^3 = 10^6 cm^3)
Mass of water = Volume of water x Density of water
Mass of water = 211,328 m^3 x 10^3 kg/m^3
Mass of water = 211,328,000 kg
Therefore, the mass of water that fell on the town is 211,328,000 kg.
To find the edge length of the cubical box containing 1.7 mol of sugar cubes, we can use Avogadro's number and the molar mass of sugar.
1. First, let's find out the number of sugar cubes in 1.7 mol. We know that 1 mol of any substance contains 6.02 x 10^23 units. Therefore, 1.7 mol will contain (1.7 mol) x (6.02 x 10^23 units/mol) = 1.024 x 10^24 sugar cubes.
2. Since we have a cubical box, the number of sugar cubes is equal to the total number of cubes inside the box. Since all edges are of equal length, we can determine this length by finding the cube root of the number of sugar cubes. In this case, the edge length is equal to the cube root of 1.024 x 10^24.
3. Using a calculator, we can find the cube root of 1.024 x 10^24, which is approximately 1.013 cm.
Therefore, the edge length of the cubical box containing 1.7 mol of sugar cubes is approximately 1.013 cm.
Now let's move on to the second question:
To find the mass of water that fell on the town, we need to calculate the volume of water first, and then multiply it by the density of water.
1. We are given that the town has an area of 32 km^2 and that it received 2.6 inches of rain in 30 minutes. To calculate the volume of water, we need to convert the area and rain amount into a common unit of measurement.
2. Firstly, convert the area of the town from km^2 to m^2. Since 1 km = 1000 m, the area of 32 km^2 is equal to 32 x (1000 m)^2 = 32 x 10^6 m^2.
3. Secondly, convert the amount of rain from inches to meters. Since 1 inch = 0.0254 m, 2.6 inches of rain is equal to 2.6 x 0.0254 m = 0.06604 m.
4. The volume of water is calculated by multiplying the area of the town by the depth of the rain. Therefore, the volume of water is (32 x 10^6 m^2) x (0.06604 m) = 2.112 x 10^6 m^3.
5. Finally, to find the mass of water, we need to multiply the volume by the density of water. The density of water is 1000 kg/m^3.
Mass of water = (2.112 x 10^6 m^3) x (1000 kg/m^3) = 2.112 x 10^9 kg.
Therefore, the mass of water that fell on the town is approximately 2.112 x 10^9 kg.
data.
-sugar cube edge of length = 1.0cm
-moles of sugar cubes =1.7moles
-edge length = unknown
-one mole =6.02 × 10^23 (consitant)
solution
let length of the edgebox =Y
hence,from volume of cubical box,V= 1.7mole of sugar cube × 6.02 × 10^23 × 1cm
V = 1.0234 × 10^24cm^3
from V = Y^3
Y^3 = 10234 × 10^20
=2.17 × 10^21 Km