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LenaWriter [7]
3 years ago
15

In 4.0 years , 40.0 kilograms of elemnt a decays to 5.0 kilograms

Chemistry
1 answer:
alexira [117]3 years ago
3 0
This problem is requiring us to find the half life of the element. We can solve this problem by using the equation as shown: A=Aoe^-(ln2/t)T where A is the final amount (5 kilogram), Ao is the initial amount (40.0 kilogram), t is the half-life (required) and lastly T time elapsed (4.0 years). Plug in the values in the equation and you should get 1.33 years as the half life. 

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Can someone please definition of atomic radius using , nucleus, valence electrons and energy.❤️
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Answer:

Explanation:

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3 years ago
30 Pointss<br> Please make sure it is correct please
Vsevolod [243]

Answer:

Option C is correct.

Explanation:

The graph shows birth and death rate in Japan. Birth rate has declined with the passage of time, as shown in the form of blue line whereas death rate is nearly constant, which can be seen as red line. More precisely, birth rate has decreased from ~30 to 10 whereas death rate is nearly ~10 all the times, although it was slightly decreased in the middle. Therefore, option C is the correct option...

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3 years ago
2) 2KClO3 --&gt; 2KCl + 3O2
aleksandrvk [35]

2 \text{ KClO}_3 \to 2 \text{ KCl}+3\text{ O}_2

a)

2 \text{ mols of KClO}_3 \equiv 3  \text{ mols of O}_2

19 \text{ mols of KClO}_3 \equiv 3\cdot 9,5  \text{ mols of O}_2

\boxed{19 \text{ mols of KClO}_3 \equiv 28,5  \text{ mols of O}_2}

b)

2 \text{ mols of KClO}_3 \equiv 2  \text{ mols of KCl}

62 \text{ mol of KClO}_3 \equiv 62  \text{ mol of KCl}

Using the atomic mass given in the periodic table:

62\cdot(39+35,5+16\cdot3) \text{ g of KClO}_3 \equiv 62  \text{ mol of KCl}

62\cdot122,5 \text{ g of KClO}_3 \equiv 62  \text{ mol of KCl}

7595 \text{ g of KClO}_3 \equiv 62  \text{ mol of KCl}

\boxed{7,595 \text{ kg of KClO}_3 \equiv 62  \text{ mol of KCl}}

c)

2 \text{ KCl}+3\text{ O}_2\to 2 \text{ KClO}_3

3 \text{ mols of O}_2 \equiv 2 \text{ mols of KCl}

Using the atomic mass given in the periodic table:

3\cdot(2\cdot 16) \text{ g of O}_2 \equiv 2\cdot(39+35,5)  \text{ g of KCl}

96\text{ g of O}_2 \equiv 149\text{ g of KCl}

\dfrac{39}{149}\cdot 96\text{ g of O}_2 \equiv \dfrac{39}{149}\cdot 149\text{ g of KCl}

\boxed{25,13\text{ g of O}_2 \equiv 39\text{ g of KCl}}

This result is an aproximation.

8 0
3 years ago
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Answer:

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4 0
2 years ago
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