The isotope samarium-151 decays into europium-151, with a half-life of approximately 96.6 years. A rock contains 5 grams of samarium-151 when it reaches its closure temperature, and it contains 0.625 grams when it is discovered.

Calculate the following:

1. The time since the rock reached its closure temperature is _ years.

2. When the rock was discovered, it had _ grams of europium-151.

The half-life of an isotope is the time it takes for half of the sample to decay. Since the decay of samarium-151 to europium-151 has a half-life of 96.6 years, we calculate how many half-lives have passed using the initial and final amount of samarium-151 in the rock. In your given problem, the amount of samarium-151 in the rock reduces from 5 grams to 0.625 grams, which means it has gone through three half-lives (since 5 grams halved three times is approximately 0.625 grams). Therefore, the time since the rock reached its closure temperature is 96.6 years multiplied by 3, which is 289.8 years.

When the rock was discovered, it had only 0.625 grams of samarium-151 left, means the remainder of the original 5 grams has become europium-151. Given that only 0.625 grams of the original 5 grams remain as samarium-151, the rest, which is 4.375 grams, would have decayed into europium-151.

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