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Oxana [17]
2 years ago
11

How much time would it take for 5.2 x 10^5 atoms of fermium-253 (half-life 3 days) to decay to 6.5 x 104 atoms?

Chemistry
1 answer:
Jlenok [28]2 years ago
4 0

Answer:

9 days, or 3 half-lives

Explanation:

5.2x10^5=520000

6.5x10^4=65000

65000/520000=1/8, or 3 half-lives

3x3=9 days

I'm not the greatest at Chem but this seems more like math than Chem :)

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What is the symbol of the ion with 36 electrons and a +2 charge?
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Answer:

Strontium

Explanation:

The atomic number of strontium is 38.

It has 38 electrons.

It is alkaline earth metal. It has two valance electrons.

Strontium loses its two electrons and form cation with +2 charge.

Electronic configuration;

Sr₃₈ = [Kr] 5s²

The valance electrons present in 5s are lost by strontium atom and form Sr⁺² cation.

it is yellowish-white metal.

It is highly reactive.

It form salt with halogens.e.g

Sr    +   Br₂    →     SrBr₂

IT react with oxygen and form oxide.

2Sr   +   O₂   →    2SrO

this oxide form hydroxide when react with water,

SrO  + H₂O   →  Sr(OH)₂

With nitrogen it produced nitride,

3Sr + N₂     →  Sr₃N₂

With acid like HCl,

Sr + 2HCl  →  SrCl₂ + H₂

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Be sure to answer all parts. A 0.365−mol sample of HX is dissolved in enough H2O to form 835.0 mL of solution. If the pH of the
Marta_Voda [28]

Answer:

The Ka is 9.11 *10^-8

Explanation:

<u>Step 1: </u>Data given

Moles of HX = 0.365

Volume of the solution = 835.0 mL = 0.835 L

pH of the solution = 3.70

<u>Step 2:</u> Calculate molarity of HX

Molarity HX = moles HX / volume solution

Molarity HX = 0.365 mol / 0.835 L

Molarity HX = 0.437 M

<u />

<u>Step 3:</u> ICE-chart

[H+] = [H3O+] = 10^-3.70 = 1.995 *10^-4

Initial concentration of HX = 0.437 M

Initial concentration of X- and H3O+ = 0M

Since the mole ratio is 1:1; there will react x M

The concentration at the equilibrium is:

[HX] = (0.437 - x)M

[X-] = x M

[H3O+] = 1.995*10^-4 M

Since 0+x = 1.995*10^-4   ⇒ x=1.995*10^-4

[HX] = 0.437 - 1.995*10^-4 ≈ 0.437 M

[X-] = x = 1.995*10^-4 M

<u>Step 4: </u>Calculate Ka

Ka = [X-]*[H3O+] / [HX]

Ka = ((1.995*10^-4)²)/ 0.437

Ka = 9.11 *10^-8

The Ka is 9.11 *10^-8

4 0
3 years ago
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