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3 years ago

4. DBearded waste of Co-60 must be stored until it is no longer radioactive. Cobalt-60

Chemistry

1 answer:

Bingel [31]3 years ago

8 0

464 g radioisotope was present when the sample was put in storage

<h3>Further explanation</h3>

Given

Sample waste of Co-60 = 14.5 g

26.5 years in storage

Required

Initial sample

Solution

General formulas used in decay:

$\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}$

t = duration of decay

t 1/2 = half-life

N₀ = the number of initial radioactive atoms

Nt = the number of radioactive atoms left after decaying during T time

Half-life of Co-60 = 5.3 years

Input the value :

$\tt 14.5=No.\dfrac{1}{2}^{26.5/5.3}\\\\14.5=No.\dfrac{1}{2}^5\\\\No=\boxed{\bold{464~g}}$

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Answer:

3.41 x10⁶ torr

Explanation:

To solve this problem we need to remember the equivalency:

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Then we can proceed to convert 4.55×10⁸ Pa into torr. To do that we just need to multiply that value by a fraction number, putting the unit that we want to convert from in the denominator, and the value we want to convert to in the numerator:

4.55x10⁸ Pa * $\frac{1torr}{133.322Pa} =$ 3.41 x10⁶ torr

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4 years ago

Why do scientists sometimes discard theories?

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4 years ago

AGO H2SO4 (aq) Al(SO4)3(aq) H28) fe Cl218) FeCl38

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Answer:

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Explanation:

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2. If 3 is now behind one H2, it must be behind the other.

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3. Al2 (SO4) 3 has 2 ahead of Al which means there will be 2Al in the reactants.

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3 years ago

Given the two reactions H2S(aq)⇌HS−(aq)+H+(aq), K1 = 9.57×10−8, and HS−(aq)⇌S2−(aq)+H+(aq), K2 = 1.46×10−19, what is the equilib

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Answer: The value of $K_c$ for the final reaction is $7.16\times 10^{25}$

Explanation:

The given chemical equations follows:

Equation 1: $H_2S(aq.)\rightleftharpoons HS^-(aq.)+H^(aq.);K_1$

Equation 2: $HS^-(aq.)\rightleftharpoons S^{2-}(aq.)+H^(aq.);K_2$

The net equation follows:

$S^{2-}(aq.)+2H^+(aq.)\rightleftharpoons H_2S(aq.);K_c$

As, the net reaction is the result of the addition of reverse of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of inverse of first equilibrium constant and the inverse of second equilibrium constant.

The value of equilibrium constant for net reaction is:

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