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kow [346]
3 years ago
9

Which describes radioactive decay of a substance? More of the radioactivity is lost during the first half-life than in later hal

f-lives. More of the radioactivity is lost during the fourth half-life than in the first half-life. Isotopes are the most stable during the first half-life. Isotopes are the least stable during the later half-lives.
Chemistry
2 answers:
Alja [10]3 years ago
8 0

Answer:

Yes, A.) "More of the radioactivity is lost during the first half-life than in later half-lives."  is the correct answer!!

Explanation:

I took the practice test on edge

vovangra [49]3 years ago
5 0

Answer:More of the radioactivity is lost during the first half-life than in later half-lives.

Explanation: Passed test with 98

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Which of the following would release the most heat? Assume the same mass of in each case. Specific heats of ice, liquid water, a
lesya692 [45]

Answer:

The process which releases most heat is E)

Explanation:

As we know that water freezes at 0ºC and vaporizes at 100ºC, we calculate the heat as follows:

  • Processes with temperatures < 0ºC : by using specific heat of ice (Sh ice) multiplied by the change in temperature (ΔT= Final Temperature - Initial Temperature)⇒ Sh ice x ΔT
  • Processes of ice melting (at 0ºC): by using heat of fusion of ice (ΔH fus) multiplied by a conversor factor (1 mol H20= 18 g)⇒ ΔHfus x 1mol/18g
  • Processes between 0ºC and 100ºC: by using specific heat of liquid water (Sh liq) multiplied by change in temperature ⇒ Sh liq x ΔT
  • Processes of water evaporation (at 100ºC): by using heat of vaporization (ΔH vap) multiplied by the conversor factor ⇒ ΔH vap x 1mol/18 g
  • Processes at a temperature >100ºC: by using specific heat of water vapor (Sh vap) multiplied by the change in temperature ⇒ Sh vap x ΔT

A) Water at -25ºC is ice. Ice is heated from -25ºC to 0ºC, then it melts at 0ºC (ice became liquid water) and then liquid water is heated from 0ºC to 70ºC. T

This is the only process in with the heat is absorbed (not releases), so it cannot be the right answer, but we calculate the heat involved to practice:

Heat= (Sh ice x ΔT) + (ΔH fus x 1/18 g) + Sh liq x ΔT

Heat= (2.05 J/g ºC x (0ºC -(-25ºC) ) + (6.01 x 10³ J/mol x 1 mol/18 g) + (4.18 J/g ºC x (70ºC-0ºC)

Heat= 51.25 J + 333,8 J +292.6 J

Heat= 677.65 J (heat is absorbed)

B) Water is cooled from 13ºC to 0ºC, then it is freezed at 0ºC and then the ice is cooled from 0ºC to -2.6 ºC

Heat= (Sh liq x ΔT) + (-ΔH melt x 1/18 g) + (Sh ice x ΔT)

Heat= 4.18 J/ºC x (0ºC- 13ºC) + (-6.01 x 10³ J/mol x 1mol/18 g) + (2.05 J/ºC x (-2.5ºc - 0ºC)

Heat= -54.34 J - 333.8 J + 5.33 J

Heat= -393.47 J (heat is released)

C) Liquid water is cooled from 74ºC to 95ºC

Heat= Sh liq x ΔT

Heat= 4.18 J/ºC x (74ºC - 95ºC)

Heat = -87.78 J (heat is released)

D) Water at 140ºC is in vapor state. Vapor at 140ºC is cooled to 110ºC (still vapor).

Heat = Sh vap x ΔT

Heat= 2.01 J/ºC x (110ºC - 140ºC)

Heat= -60.3 J (heat is released)

E) Vapor at 106ºC is cooled to 100ºC, then it condenses at 100ºC (convertion from gas to liquid), and liquid water is cooled from 100ºC to 88ºC.

Heat= (Sh vap x ΔT) + (-ΔHvap x 1mol/18g) + (Sh liq x ΔT)

Heat= (2.01 J/ºC x (100ºC-106ºC)) - (40.7 x 10³ J/mol x 1mol/18 g) + (4.18 J/ºC x (88ºC -100ºC)

Heat= -2323.32 J (heat is released) <u>THIS IS THE RIGHT ANSWER</u> (the more negative= the more released)

7 0
3 years ago
In terms of bonds what would the molecule C6H12 be classified as
docker41 [41]

It has a double C=C bond so that means it's unsaturated, but it can also be a cyclic compound with only simple C-C bonds

5 0
3 years ago
A student wants to see how temperature affects the
Musya8 [376]

Answer:

To monitor Pressure increase

Explanation:

the more pressure, the higher the production of the gas, which can be recorded, more pressure meaning higher production. Then vary the temperature.

5 0
3 years ago
Read 2 more answers
How many molecules Of H20 are<br> equivalent to 97.2 g H20?<br> (H = 1.008 g/mol, O = 16.00 g/mol)
AleksandrR [38]

Answer:

m(H₂O) = 97,2 g.n(H₂O) = m(H₂O) ÷ M(H₂O).n(H₂O) = 97,2 g ÷ 18

Explanation:

8 0
2 years ago
How many grams of P4O10 would be produced when 0.700 mole of phosphorus is burned?
AysviL [449]
We can solve the equation and show the solution below:

Oxygen atomic number is 16.
Phosphorus atomic number is 32.

We have the molecular weight:
Molecular weight = (31*4) + (16*10)
Molecular weight = 284 grams/mol

Solving for the grams:

0.4 mole (for P4) * (1 mol P4O10/1 mol P4) * (284 grams P4O10/1 mole P4O10)
Total grams = 113.6

The answer is 113.6 grams.
8 0
3 years ago
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