Use the table of data for a decomposition reaction to answer the question Concentration (C) at T1 (g/L) C at T2 (g/L) C at T3 (g
/L) Cat T4 (g/L) 100 300 50 50 Reactant 100 0 125 125 Product 1 100 125 125 0 Product N In a decomposition reaction a reactant breaks down to form two products. At which time did the system first reach chemical equilibrium? (1 point)
1 answer:
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1. <em>Balancing nuclear equations </em>
Answer:
Explanation:
The main point to remember in balancing nuclear equations is that the <em>sums of the superscripts</em> (the mass numbers) and the <em>subscripts</em> (the nuclear charges) <em>must balance</em>.
Mass numbers: 37 + 0 = 37; balanced.
Charges: 18 + 1 = 17; balanced
B is <em>wrong</em>. Mass numbers not balanced. 6 +2(1) ≠ 4 + 3.
C is <em>wron</em>g. Mass numbers not balanced. 254 + 4 ≠ 258 + 2(1).
D is <em>wron</em>g. Mass numbers not balanced. 14 + 4 ≠ 17 + 2.
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2. <em>Amount remaining </em>
Answer:
D. 5.25 g
Explanation:
The half-life of Th-234 (24 da) is the time it takes for half the Th 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/(da</u>) <u>remaining</u> <u>remaining/g</u>
1 24 ½ 21.0
2 48 ¼ 10.5
3 72 ⅛ 5.25
4 96 ¹/₁₆ 2.62
We see that 72 da is three half-lives, and the amount of Th-234 remaining is 5.25 g.
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3.<em> Calculating the half-life </em>
Answer:
a. 2.6 min
Explanation:
The fraction of the original mass remaining is 1.0 g/4.0 g ≈ ¼.
We saw from the previous table that it takes two half-lives to decay to ¼ of the original amount.
2 half-lives = 5.2 min Divide both sides by 2
1 half-life = 5.2 min/2 = 2.6 min