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lesya [120]
3 years ago
7

Can you answer these two questions right please and thank you

Chemistry
1 answer:
lisov135 [29]3 years ago
6 0

Answer:

B. LZAQD

A. Younger than A but older than Q

Explanation:

To solve this problem, we simply apply the stratigraphic laws which are the law of superposition and principle of cross cutting.

  • According to the law of superposition, in an undeformed sequence of strata, the oldest layer is always at the bottom and the youngest on top.
  • In this case, we have a little disturbance but it did not affect much of the original bedding.
  • So, the rock ages from L to Z to A to Q and D. L is the oldest and D is the youngest.

According to the principle of cross-cutting "features that cuts through a rock are younger than the layers they cut through". In this problem, the fault cuts through layers LZ and A which suggests that these layers are older than the faulting event. Layer Q is unaffected by the faulting so, the fault is older than the layer.

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The 131 I isotope emits radiation and particles and has an 8-day half-life. Orally administered, it concentrates in the thyroid, where the thyroid gland is destroyed by the particles.

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Half of the iodine-131 will still be present after 8.1 days.

The amount of iodine-131 will again be halved after 8.1 additional days, for a total of 8.1+8.1=16.2 days, reaching (1/2)(1/2)=1/4 of the initial amount.

The quantity of iodine-131 will again be halved after 8.1 more days, for a total of 16.2+8.1+8.1=32.4 days, to (1/4)(1/2)(1/2)=1/16 of the initial quantity.

If the original dose of iodine-131 was 50mg, the residual dose will be (50mg)*(1/16)=3.124mg after 32.4 days.

Learn more about the Half life of radioactie element with the help of the given link:

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