Answer:
2.10 × 10⁹ yr
Step-by-step explanation:
The half-life of U-235 is the time it takes for half the U to decay.
After one half-life, half (50 %) of the original amount will remain.
After a second half-life, half of that amount (25 %) will remain, and so on.
We can construct a table as follows:
No. of Fraction Amount
<u>half-lives</u> <u>t/(yr × 10⁶)</u> <u>remaining</u> <u>remaining/g </u>
1 700 ½ 10.0
2 1400 ¼ 5.00
3 2100 ⅛ 2.50
4 2800 ¹/₁₆ 1.25
We see that 2100 × 10⁶ yr is three half-lives, and the amount of U-235 remaining is 2.50 g.
It takes 2.10× 10⁹ yr for the U-238 to decay to 2.50 g.
Oxygen gas produced : 0.7 g
<h3>Further explanation</h3>
Given
10.0 grams HgO
9.3 grams Hg
Required
Oxygen gas produced
Solution
Reaction⇒Decomposition
2HgO(s)⇒2Hg(l)+O₂(g)
Conservation of mass applies to a closed system, where the masses before and after the reaction are the same
mass of reactants = mass of products
mass HgO = mass Hg + mass O₂
10 g = 9.3 g + mass O₂
mass O₂ = 0.7 g
48.024 grams is the correct mass, because helium has an atomic mass of 4.002g, so all you have do is multiply 12 by 4.002 to get your mass of 12 helium atoms.
Im kinda puzzled on that particular question
Sugar dissolving in warm water because when the sugar dissolves you can get the sugar back by letting the water evaporate.,