Answer:
V₂ = 0.62 L
Explanation:
Given data:
Initial volume = 2.4 L
Initial temperature = 25°C
Final temperature = -196°C
Final volume = ?
Solution:
Initial temperature = 25°C (25+273 = 298 K)
Final temperature = -196°C ( -196+273 = 77 K)
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 2.4 L × 77 K / 298 k
V₂ = 184.8 L.K / 298 K
V₂ = 0.62 L
The final temperature in Celsius of the metal block is 49°C.
<h3>How to find the number of moles ?</h3>
Moles water = 
= 
= 0.0266 moles
Heat lost by water = 0.0266 mol x 44.0 kJ/mol
= 1.17 kJ
= 1170 J [1 kJ = 1000 J]
Heat lost = Heat gained
Heat gained by aluminum = 1170 J
1170 = 55 x 0.903 (T - 25) = 49.7 T - 1242
1170 + 1242 = 49.7 T
T = 48.5°C (49°C at two significant figures)
Thus from the above conclusion we can say that The final temperature in Celsius of the metal block is 49°C.
Learn more about the Moles here: brainly.com/question/15356425
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Answer:
tritium and deuterium are combined and result in the formation of helium
The question is incomplete. The complete question is:
The half-life for the decay of carbon-14 is 5.73x10^3 years. Suppose the activity due to the radioactive decay of the carbon-14 in a tiny sample of an artifact made of woodfrom an archeological dig is measured to be 2.8x10^3 Bq. The activity in a similiar-sized sample of fresh wood is measured to be 3.0x10^3 Bq. Calculate the age of the artifact. Round your answer to 2 significant digits.
Answer:
570 years
Explanation:
The activity of the fresh sample is taken as the initial activity of the wood sample while the activity measured at a time t is the present activity of the wood artifact. The time taken for the wood to attain its current activity can be calculated from the formula shown in the image attached. The activity at a time t must always be less than the activity of a fresh wood sample. Detailed solution is found in the image attached.