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Gnom [1K]
3 years ago
14

Thulium-167 has a half-life of 9.25 days. If you begin with 48 grams of thulium-167, how much of the original isotope will remai

n after 37 days?
Chemistry
1 answer:
fgiga [73]3 years ago
4 0

Answer:

2.3125g

Explanation:

Half-life referred to the time required for a quantity to reduce to half of its initial value, It used to calculate how unstable atoms undergo, or the period of time and atom can survive, radioactive decay.

Given:

t(1/2)= 9.25days

Initial mass of Thulium-167 = 48grams

We need to calculate the remaining amount after 37days.

Since we know that 1 half life = 9.25 days

Then 37 days means ( 37/9.25) half lives

37days means 4 half life

That means the 38grams of Thulium-167 will be halved by 4 times.

Then the ratio between the initial Amount and the amount remaining after 37 days can be calculated as. 0.5^(4)

= 37days × 0.5^(4)

= 2.3125g

the remaining amount of Thallium-167 after 37days is 2.3125g

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4 0
3 years ago
24 g of magnesium were burned in oxygen. The compound formed had a mass of 40 g. Explain why the mass had gone up.
DIA [1.3K]

Answer :

According to the law of conservation of mass, the mass of reactants must be equal to the mass of products.

The balanced chemical reaction is,

Mg+\frac{1}{2}O_2\rightarrow MgO

As we know that the molar mass of magnesium is 24 g/mole, the molar mass of O_2 is 32 g/mole and the molar mass of magnesium oxide is 40 g/mole.

From the given balanced reaction, we conclude that

As, 1 mole of magnesium react \frac{1}{2} mole of oxygen to give 1 mole of magnesium oxide.

So, the mass of Mg is 24 g, the mass of O_2=\frac{1}{2}\times 32=16g and the mass of MgO is 40 g.

That means 24 g of Mg react with 16 g O_2 to give 40 g of MgO.

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