The activity of the sample when it was shipped from the manufacturer is 4.54 mCi
<h3>How to determine the number of half-lives that has elapsed </h3>
From the question given above, the following data were obtained:
- Time (t) = 48 hours
- Half-life (t½) = 14.28 days = 14.28 × 24 = 342.72 hours
- Number of half-lives (n) =?
n = t / t½
n = 48 / 342.72
n = 0.14
<h3>How to determine the activity of the sample during shipping </h3>
- Number of half-lives (n) = 0.14
- Original activity (N₀) = 5.0 mCi
- Activity remaining (N) =?
N = N₀ / 2ⁿ
N = 5 / 2^0.14
N = 4.54 mCi
Thus, the activity of the sample during shipping is 4.54 mCi
Learn more about half life:
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<h3><u>Answer;</u></h3>
exceeds evaporation over land
Precipitation<u> exceeds evaporation over land </u>
<h3><u>Explanation;</u></h3>
- <em><u>In order to maintain earths water balance, evaporation exceeds precipitation over oceans but precipitation exceeds evaporation over land.</u></em>
- Water evaporates into the atmosphere from the ocean and to a much lesser extent from the continents. Winds transport this moisture-laden air, often great distances, until conditions cause the moisture to condense into clouds and to precipitate and fall.
- Most precipitation originates by evaporation from the oceans. Over time, water evaporated from the oceans is replenished by inflow of freshwater from rivers and streams.
To calculate the amount of heat transferred when an amount of reactant is decomposed, we must look at the balanced reaction and its corresponding heat of reaction. In this case, we can see that 252.8 kJ of heat is transferred per 2 moles of CH3OH used. When 22 g of CH3OH is used, 86.9 kJ is absorbed.
Answer:
Q = 60192 j
Explanation:
Given data:
Volume of water = 0.45 L
Initial temperature = 23°C
Final temperature = 55°C
Amount of heat absorbed = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 55°C - 23°C
ΔT = 32°C
one L = 1000 g
0.45 × 1000 = 450 g
Specific heat capacity of water is 4.18 j/g°C
Q = m.c. ΔT
Q = 450 g. 4.18 j/g°C. 32°C
Q = 60192 j
Answer:
44
Explanation:
Given that :
Mass of solute = Mass of urea = 16g
Mass of water = 20g
Mass of solution = (mass of solute + mass of solvent) = (mass of urea + mass of water) = (16g + 20g) = 36g
Percentage Mass = (mass of solute / mass of solution) * 100%
Percentage Mass = (16 / 36) * 100%
Percentage Mass = 0.4444444 x 100%
Percentage Mass = 44.44%
Percentage Mass = 44%
