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Liono4ka [1.6K]
3 years ago
5

A sample of helium has a volume of 5 liters and a pressure of 699 mmHg. If the final volume is 5.7 Liters, its final temperature

303 K, and the final pressure 800 mmHg, what was the initial temperature of helium?
Chemistry
1 answer:
malfutka [58]3 years ago
6 0

Answer:

The initial temperature of helium was T1 = 232.23 K

Explanation:

Given data:

Initial volume V1 = 5 L

Initial pressure P1 = 699 mmHg

Final pressure P2 = 800 mmHg

Final volume V2 = 5.7 L

Final temperature T2= 303 K

Initial temperature T1 = ?

Solution:

Formula:

P1V1/T1 = P2V2/T2

T1 = T2 × P1V1/P2V2

T1 = 303 K × 699 mmHg × 5 L / 800 mmHg × 5.7 L

T1 = 1058985/ 4560

T1 = 232.23 K

initial temperature of helium was 232.23 k.

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The half-life of tritium (H-3) is 12.3 years. If 48.0mg of tritium is released from a nuclear power plant during the course of a
Rudiy27

Answer:

The amount left after 49.2 years is 3mg.

Explanation:

Given data:

Half life of tritium = 12.3 years

Total mass pf tritium = 48.0 mg

Mass remain after 49.2 years = ?

Solution:

First of all we will calculate the number of half lives.

Number of half lives = T elapsed/ half life

Number of half lives =  49.2 years /12.3 years

Number of half lives =  4

Now we will calculate the amount left after 49.2 years.

At time zero 48.0 mg

At first half life = 48.0mg/2 = 24 mg

At second half life = 24mg/2 = 12 mg

At 3rd half life = 12 mg/2 = 6 mg

At 4th half life =  6mg/2 = 3mg

The amount left after 49.2 years is 3mg.

6 0
4 years ago
What is the specific heat of a 123 g substance that requires 4.56 J of heat in
vladimir1956 [14]

Answer:  A. 0.00301J/g^0C

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

Q=m\times c\times \Delta T

Q = Heat absorbed  = 4.56 J

m = mass of substance = 123 g

c = specific heat capacity = ?

Change in temperature ,\Delta T=T_f-T_i=12.32^0C

Putting in the values, we get:

4.56J=123g\times c\times 12.32^0C

c=0.00301J/g^0C

The specific heat of a 123 g substance that requires 4.56 J of heat in  order to increase its temperature by 12.32 °C is 0.00301J/g^0C

8 0
3 years ago
A solution of pH 3 was mixed with a solution of pH 9. What is the final pH could be ?
chubhunter [2.5K]

The solution of which contains of PH 3 called acidic medium which is more stronger acid called HCl when reacts to the stronger base base NaOH forms the acid base reaction .

According to acid base equation you consider ph5 as PH 9 if you understand it then tell me below

3 0
3 years ago
What is the answer for this question
cestrela7 [59]

Answer:

The answer to your question is the second choice

Explanation:

Formula to calculate the heat of any substance

             Q cal= mCcal ΔTcal

Formula to calculate the heat of a metal. The heat will be negative because it releases heat.

                  Qm = -mCmΔTm

Now equal both formula

                   Qcal = Q m

           mCcalΔTm = -mCmetΔTmet

-Solve for Cmet

                 Cmet = -[mCcalΔTm] / mΔTmet

6 0
3 years ago
In which part of the cell is the majority of the energy released from the breakdown of glucose
Vladimir [108]

<em>Answer</em><em>:</em>

<em>Glycolysis</em>

<em>E</em><em>xplanation</em><em> </em><em>:</em>

Glycolysis is the first step in the breakdown of glucose to extract energy for cell metabolism.Many living organisms carry out glycolysis as part of their metabolism. Glycolysis takes place in the cytoplasm of most prokaryotic and all eukaryotic cells.

7 0
3 years ago
Read 2 more answers
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