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kow [346]
3 years ago
9

Which describes radioactive decay of a substance? More of the radioactivity is lost during the first half-life than in later hal

f-lives. More of the radioactivity is lost during the fourth half-life than in the first half-life. Isotopes are the most stable during the first half-life. Isotopes are the least stable during the later half-lives.
Chemistry
2 answers:
Alja [10]3 years ago
8 0

Answer:

Yes, A.) "More of the radioactivity is lost during the first half-life than in later half-lives."  is the correct answer!!

Explanation:

I took the practice test on edge

vovangra [49]3 years ago
5 0

Answer:More of the radioactivity is lost during the first half-life than in later half-lives.

Explanation: Passed test with 98

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Four properties of water are listed.
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It is colorless and oderless is a physical property by telling what color.
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3 years ago
Plutonium−239 (t1/2 = 2.41 × 104 yr) represents a serious nuclear waste hazard. if seven half-lives are required to reach a tole
dimulka [17.4K]
The half-life of Plutonium−239, t1/2 is 2.41 × 10⁴<span> yrs

time taken to reach tolerable level = seven half-lives
                                                      = 7 x t1/2
                                                      = 7 x </span>2.41 × 10⁴ yrs
                                                      = 168700 yrs
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Hence, the period of time that <span>Plutonium-239 must be stored is </span>1.687 x 10⁵ years.
5 0
3 years ago
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Brrunno [24]
A. immigration, how is this chemistry?
8 0
3 years ago
What is the current produced when a 12-Volt battery encounters a resistance of 20 Ohms?
Verdich [7]

Answer:

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4 0
3 years ago
On the basis of the information above, a buffer with a pH = 9 can best be made by using
telo118 [61]

Answer:

D H2PO4– + HPO42–

Explanation:

The acid dissociation constant for \mathbf{H_3PO_4 , H_2PO^{-}_4 ,  HPO_4^{2-}} are \mathbf{7\times 10^{-3}, \ \ 8\times 10^{-8} ,\ \  5\times 10^{-13}} respectively.

\mathbf{pka (H_3PO_4) = -log (7\times 10^{-3} )=2.2}

\mathbf{pka (H_2PO_4^-) = -log (8\times 10^{-8} )=7.1}

\mathbf{pka (HPO_4^{2-}) = -log (5\times 10^{-13} )=12.3}

The reason while option D is the best answer is that, the value of pKa for both

\mathbf{H_2PO^{-}_4 ,\  \& \  HPO_4^{2-}} lies on either side of the desired pH of the buffer. This implies that one is slightly over and the other is slightly under.

Using Henderson-Hasselbach equation:

\mathbf{pH = pKa + log \Big( \dfrac{HPO_4^{2-}}{H_2PO_4^-} \Big)}

3 0
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