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iren2701 [21]
11 months ago
5

The half life of plutonium-239 (239pu, pu-239) is 24,100 years. How much plutonium will remain after 1000 years if the initial a

mount is 5 g?.
Chemistry
1 answer:
Anestetic [448]11 months ago
8 0

The mass of plutonium that will remain after 1000 years if the initial amount is 5 g when the half life of plutonium-239 (239pu, pu-239) is 24,100 years is 2.5 g

The equation is Mr=Mi(1/2)^n

where n is the number of half-lives

Mr is the mass remaining after n half lives

Mi is the initial mass of the sample

To find n, the number of half-lives, divide the total time 1000 by the time of the half-life(24,100)

n=1000/24100=0.0414

So Mr=5x(1/2)^1=2.5 g

The mass remaining is 2.5 g

  • The half life is the time in which the concentration of a substance decreases to half of the initial value.

Learn more about half life at:

brainly.com/question/24710827

#SPJ4

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Why is a mass movement of mud called a flow?
wariber [46]
In terms of a deeper scientific reason, I am not sure, but the basic reason is quite simple. "Mud" tends to look like a mix between a solid, dirt, and a liquid, water or some other liquid. Since it is, in fact, a cross between a solid and a liquid, it has properties of both. It has certain physical and visual properties that only a solid would have, such as texture and opaqueness, but it also has physical properties of a liquid. Since it leans more towards the liquid side than the solid side, we say mud "flows" rather than saying that it "rolls" or "bounces".
5 0
3 years ago
What will happen if a candle will be covered with an inverted jar filled will nitrogen
exis [7]

Answer:

The flame will burn out

Explanation:

The oxygen will be kept out of the jar, without oxygen the flame cannot last long. The flame will eventually burn out. Nitrogen gas has no role to play in the burning. Nitrogen slows down the burning rate.

8 0
2 years ago
g A piece of solid Zn metal is put into an aqueous solution of Cu(NO3)2. Write the net ionic equation for any single-replacement
Nimfa-mama [501]

Answer:

Zn(s) + Cu²⁺(aq) ⇒ Zn²⁺(aq) + Cu(s)

Explanation:

Let's consider the molecular single displacement equation between Zn and Cu(NO₃)₂

Zn(s) + Cu(NO₃)₂(aq) ⇒ Zn(NO₃)₂(aq) + Cu(s)

The complete ionic equation includes all the ions and insoluble species.

Zn(s) + Cu²⁺(aq) + 2 NO₃⁻(aq) ⇒ Zn²⁺(aq) + 2 NO₃⁻(aq) + Cu(s)

The net ionic equation includes only the ions that participate in the reaction and insoluble species.

Zn(s) + Cu²⁺(aq) ⇒ Zn²⁺(aq) + Cu(s)

4 0
3 years ago
Be sure to answer all parts. Find the pH during the titration of 20.00 mL of 0.1000 M triethylamine, (CH3CH2)3N (Kb = 5.2 × 10−4
Nataliya [291]

Answer:

(a) 10.62

(b) 2.82

(c) 1.95

Explanation:

The neutralization reaction in this question is

(CH3CH2)3N + HCl   ⇒ (CH3CH2)3NH⁺ + Cl⁻

The problem  can be  solved by calculating the number of moles of triethylamine after  addition of the portions of HCl. Since it is a weak base if it is not consumed completely, that is in excess we will have a buffer of a waek base. If its consumed completely the pH will be determined by the strong acid HCl.

The pOH for a buffer of a weak base is gven by

pOH = pKb + log [(CH3CH2)3NH⁺] / [(CH3CH2)3N]

(a) 11 mL of 0.100 M HCl

mol HCl = 0.011 L x 0.100 mol/L = 0.0011 mol HCl

mol  (CH3CH2)3N reacted = 0.0011 mol

mol (CH3CH2)3NH⁺ produced = 0.0011 mol

mol (CH3CH2)3N  initially = 0.020 L x 0.1000 mol/L 0.0020 mol

mol (CH3CH2)3N left = 0.0020 mol - 0.0011 = 0.0009 mol

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mol  (CH3CH2)3N consumed = 0.0020 mol

This is so  because the acid will consume completely the 0.0020 mol of the weak base  we had originally present.

Now the problem circumscribes to that of calculating the pH of the unreacted HCl

Total Vol = 0.0206 L + 0.02 L = 0.0406 L

mol HCl = 0.0206 L x .100 = 0.00206 mol

mol HCl left = 0.00206 mol - 0.0020 mol = 0.00006 mol

[HCl] = 0.00006 mol / 0.0406 L = 0.0015 M

Since HCl is a strong acid ( 100 % ionization) :

pH = - log [H⁺] = - log ( 0.0015 ) = 2.82

(c) We will compute the pH in  the same way we did for part (b)

mol HCl = 0.025 L x 0.100 mol/L = 0.0025 mol

mol HCl left = 0.0025 mol  - 0.0020 mol = 0.0005

Total Volume = 0.020 L + 0.025 L = 0.045 L

[HCl] = 0.0005 mol / 0.045 L = 0.111

pH = - log ( 0.111) = 1.95                                            

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3 years ago
Is "Hypooxide" even a thing???
bulgar [2K]

Answer:

yes it is

Explanation:

4 0
3 years ago
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