True would be your answer
Answer:
0.25 g of U-235 isotope will left .
Formula used :
where,
N = amount of U-235 left after n-half lives = ?
= Initial amount of the U-235 = 1.00 g
n = number of half lives passed = 2
0.25 g of U-235 isotope will left .
2
2
3
4
20
Is the answer for the question
Answer:
5.37 L
Explanation:
To solve this problem we need to use the PV=nRT equation.
First we <u>calculate the amount of CO₂</u>, using the initial given conditions for P, V and T:
- P = 785 mmHg ⇒ 785/760 = 1.03 atm
- T = 18 °C ⇒ 18 + 273.16 = 291.16 K
1.03 atm * 4.80 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 291.16 K
We <u>solve for n</u>:
Then we use that value of n for another PV=nRT equation, where T=37 °C (310.16K) and P = 745 mmHg (0.98 atm).
- 0.98 atm * V = 0.207 mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 310.16 K
And we <u>solve for V</u>:
