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bulgar [2K]
3 years ago
5

Cobalt-60, with a half-life of 5.0 days, is used in cancer radiation treatments. A hospital purchases a 30.0g supply of it and s

tores it on a shelf in a closet. The next time a hospital worker grabs the bottle, though, they find that it only contains 3.75 g remaining! Assuming the bottle had never been opened, how long was the bottle sitting on the shelf before the worked grabbed it?
Chemistry
1 answer:
Rudik [331]3 years ago
8 0

Answer:

Bottle is sitting on the shelf for 15 days.

Explanation:

Given data:

Co-60 half life = 5 days

Total amount = 30.0 g

Amount left = 3.75 g

Time taken = ?

Solution:

AT time zero = 30 g

AT first half life= 30g /2 = 15 g

At 2nd half life = 15 g/ 2 = 7.5 g

At 3rd half life = 7.5 g/2 = 3.75 g

Now we will calculate the sitting time of bottle.

Half life = Time taken / number of half lives

3× 5 days = time taken

Time taken = 15 days

Bottle is sitting on the shelf for 15 days.

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Pb(NO3)2(aq) + KCl(aq) ---- → KNO3(aq) + PbC12(s)
emmasim [6.3K]
If you’re asking to balance the equation then:

Pb(NO3)2(aq) + 2KCl(aq) -> 2KNO3(aq) + PbCl2(s)

Just remember: the equations at the end is Cl not C12

Note: the small number on the bottom (subscripts) apply to the one element if it’s inside the bracket and if the small number is on the outside of the bracket it applies to all the elements. For example the 3 in (NO3)2 applied only to the O (oxygen) and the 2 applies to both N and O but don’t forget it’s multiplied. So it would be 2 N’s and 6 O’s bc the 3 multiplies with the 2 only for the O.
6 0
2 years ago
“Representative elements”? Which groups of the periodic table contain the “representative elements”?
Phoenix [80]

S- and P-Block elements are known as Representative elements.

Groups 1-2 and Groups 13-18

4 0
3 years ago
Calculate the specific heat capacity for a 15.3-g sample of gold that absorbs 87.2 J when its temperature increases from 35.0 °C
diamong [38]

Answer:

The specific heat of gold is 0.129 J/g°C

Explanation:

Step 1: Data given

Mass of gold  = 15.3 grams

Heat absorbed = 87.2 J

Initial temperature = 35.0 °C

Final temperature = 79.2 °C

Step 2:

Q = m*c*ΔT

⇒ Q =the heat absorbed = 87.2 J

⇒ m = the mass of gold = 15.3 grams

⇒ c = the specific heat of gold = TO BE DETERMINED

⇒ ΔT = The change in temperature = T2 - T1 = 79.2 - 35.0 = 44.2 °C

87.2 J = 15.3g * c * 44.2°C

c = 87.2 / (15.3 * 44.2)

c = 0.129 J/g°C

The specific heat of gold is 0.129 J/g°C

4 0
3 years ago
Aluminum has a density of 2.7 g/cm3, how much space in cm3 would 81 grams of aluminum occupy? Show steps to answering this equat
Evgesh-ka [11]

Answer:

30 cm³

Explanation:

Step 1: Given data

  • Density of aluminum (ρ): 2.7 g/cm³
  • Mass of aluminum (m): 81 g
  • Volume occupied by aluminum (V): ?

Step 2: Calculate the volume occupied by aluminum

The density of aluminum is equal to its mass divided by its volume.

ρ = m/V

V = m/ρ

V = 81 g / 2.7 g/cm³

V = 30 cm³

4 0
3 years ago
Bruh Im 4th on the leaderboard How is this possible
forsale [732]

Answer:

I guess you just answered a lot of questions

Explanation:

Thanks for the points btw :)

4 0
3 years ago
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