The radon-222 sample has a half-life of 3.8 days, and we are asked how many times would the mass divide in half after 23 days. First we calculate the amount of times division occurs by taking the number of days and dividing that by the number of days for one half-life to occur: 23/3.8 = 6.05.
We have 198.6 grams of sample, and we are going to divide it in half 6 times to determine how much of it remains after 23 days:
198.6/2 = 99.3 grams
99.3/2 = 49.65 grams
49.65/2 = 24.83 grams
24.83/2 = 12.41 grams
12.41/2 = 6.21 grams
6.21/2 = 3.1 grams
Therefore, we are left with 3.1 grams of radon-222 after 23 days if one half-life equals to 3.8 days.
The number 6.022 × 1023 indicating the number of atoms or molecules in a mole of any substance
The change in temperature had the greatest effect at changing the volume of the balloon.
<h3>What are the gas laws?</h3>
The gas laws are used to describe the parameters that has to do with gases.
Given that;
P1 = 98.5 kPa
T1 = 18oC or 291 K
V1 = 74.0 dm3
P2 = 7.0 kPa
V2 = ?
T2 = 18oC or 291 K
P1V1/T1 = P2V2/T2
P1V1T2 =P2V2T1
V2= P1V1T2/P2T1
V2 = 98.5 kPa * 74.0 dm3 * 291 K/ 7.0 kPa * 291 K
V2 = 1041.3 dm3
When;
V1 = 1041.3 dm3
T1 = 291 K
V2 = ?
T2 = 80oC or 353 K
V1/T1 = V2/T2
V1T2 = V2T1
V2 = V1T2/T1
V2 = 1041.3 dm3 * 353 K/291 K
V2 = 1263 dm3
The change in temperature had the greatest effect at changing the volume of the balloon.
Given that
V1 = 100 cm^3
T1 = 273 K
P1 = 1.01 * 10^5 Pa
V2 = ?
P2 = 3.00 x 10^-4 Pa
T2 = -180oC or 255 K
V2= P1V1T2/P2T1
V2 = 1.01 * 10^5 Pa * 100 cm^3 * 255 K / 3.00 x 10^-4 Pa * 273 K
V2 = 3.14 * 10^10 cm^3
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Answer: Boyle's Law
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Boyle's Law is the pressure-volume law and it relates pressure and volume at constant temperature. Boyle's law states that pressure and volume vary inversely, meaning that as one goes up, the other one goes down.
Answer:
This process naturally occurs in the environment, where it is carried out by specialized bacteria.
Explanation:
