The amount remaining at the end of 5 half-lives is 7.81×10¹³ g
From the question given above, the following data were obtained:
- Half-life (t½) = 5730 years
- Original amount (N₀) = 2.5×10¹⁵ g
- Number of half-lives (n) = 5
- Amount remaining (N) =?
The amount remaining can be obtained as follow:
N = 1/2ⁿ × N₀
N = 1/2⁵ × 2.5×10¹⁵
N = 1/32 × 2.5×10¹⁵
N = 0.03125 × 2.5×10¹⁵
N = 7.81×10¹³ g
Therefore, the amount remaining after 5 half-lives is 7.81×10¹³ g
Learn more about half-life: brainly.com/question/25783920
They are identical to the nucleus of the helium-4 atom. So the answer is helium atoms.
Hope this helps! :)
First convert each mass to moles by dividing by the molar mass.
0.347g Oxygen = 0.022 moles
0.260g Carbon = 0.022 moles
1.537g Chlorine = 0.043 moles
Divide each by smallest mole vale (0.022)
Oxygen = 0.022/0.022 = 1
Carbon = 0.022/0.022 = 1
Chlorine = 0.043/0.022 = 2
Therefore the empirical formula is COCl2
Answer:
7.5 mol
Step-by-step explanation:
2Al + 3Br₂ ⟶ 2AlBr₃
You want to convert moles of AlBr₃ to moles of Br₂.
The molar ratio is 3 mol Br₂:2 mol AlBr₃.
Moles of Br₂ = 5 × 3/2
Moles of Br₂ = 7.5 mol Br₂
You need 7.5 mol of Br₂ to produce 5 mol of AlBr₃.
