The total volume of the Rogers Center is 1.6 x 10^6 m cubed. If the volume of 100 peas is about 55 cm cubed, how many Rogers Centers would 6.02 x 10^23 peas fill?
100 peas/55cm^3 * (100cm/1m)^3 = 10^8/55 peas/m^3
6.02*10^23 peas / (10^8/55 peas/m^3) * 1RC/1.6*10^6m^3
= 6.02*10^23 * 55/10^8 / 1.6*10^6 = 206,937,500,000 RC
be aware of significant figures
(5.5E-7 m^3 / pea) * 6.02E23 peas / 1.6E6 m^3
To solve this problem, we need to determine the number of peas that would fill one Rogers Center and then calculate how many Rogers Centers would be required for a given number of peas.
First, let's find out how many peas are needed to fill one Rogers Center. We have the volume of a pea as 55 cm^3 and the volume of the Rogers Center as 1.6 x 10^6 m^3.
1 cm^3 is equal to (1/100)^3 = 1/1,000,000 m^3. Therefore, the volume of one pea can be converted to m^3 by dividing by 1,000,000.
Volume of one pea in m^3 = 55 cm^3 / 1,000,000 = 0.000055 m^3
Now, to find out how many peas would fill one Rogers Center, divide the volume of the Rogers Center by the volume of one pea:
Number of peas to fill one Rogers Center = Volume of Rogers Center / Volume of one pea
= 1.6 x 10^6 m^3 / 0.000055 m^3
= 2.9 x 10^10 peas
So, one Rogers Center would be filled by approximately 2.9 x 10^10 peas.
Next, we need to calculate how many Rogers Centers would be required to hold 6.02 x 10^23 (Avogadro's number) peas.
Number of Rogers Centers needed = Total number of peas / Number of peas per Rogers Center
= 6.02 x 10^23 peas / 2.9 x 10^10 peas
= 2.07 x 10^13 Rogers Centers
Therefore, 6.02 x 10^23 peas would fill approximately 2.07 x 10^13 Rogers Centers.