The number of grams of radon 222 did it have 15.2 ago was 49.6 grams( answer C)
<u>calculation</u>
- calculate the number of half life it has covered from 15.2 days to 3.8 days
that is divide 15.2/ 3.8 = 4 half life
- half life is time taken for a radio activity of a specified isotope to fall to half its original mass
therefore 3.8 days ago it was 3.1 x2 = 6.2 grams
7.6 days ago it was 6.2 x2 = 12.4 grams
11.4 days ago it was 12.4 x2= 24.8 grams
15.2 days ago it was 24.8 x2=49.6 grams
A compound is a substance that can be separated into simpler substances only by chemical means.
Nitrogen is 5 valence electrons
oxygen is 6 valence electrons
fluorine is 7 electrons
The mass of sodium sulfite that was used will be 1,890 grams.
<h3>Stoichiometric problems</h3>
First, the equation of the reaction:

The mole ratio of SO2 produced and sodium sulfite that reacted is 1:1.
Mole of 960 grams SO2 = 960/64 = 15 moles
Equivalent mole of sodium sulfite that reacted = 15 moles
Mass of 15 moles sodium sulfite = 15 x 126 = 1,890 grams
More on stoichiometric problems can be found here: brainly.com/question/14465605
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Answer:
Explanation:
<u>1) First law of thermodynamic (energy balance)</u>
- Heat released by the the hot water (345K ) = Heat absorbedby the cold water (298 K) + Heat absorbed by the calorimeter
<u>2) Energy change of each substance:</u>
Heat released or absorbed = mass × Specific heat × change in temperature
- density of water: you may take 0.997 g/ ml as an average density for the water.
- mass of water: mass = density × volume = 50.0 ml × 0.997 g/ml = 49.9 g
- Specif heat of water: 1 cal / g°C
- Heat released by the hot water:
Heat₁ = 49.9 g × 1 cal / g°C × (345 K - 317 K) = 49.9 g × 1 cal / g°C × (28K)
- Heat absorbed by the cold water:
Heat₂ = 49.9 g × 1 cal / g°C × (317 K - 298 K) = 49.9 g × 1 cal / g°C × (19K)
- Heat absorbed by the calorimeter
Heat₃ = Ccal × (317 K - 298 K) = Ccal × (19K)
<u>4) Balance</u>
49.9 g × 1 cal / g°C × (28 K) = 49.9 g × 1 cal / g°C × (19 K) + Ccal × (19 K)
Ccal = [49.9 g × 1 cal / g°C × (28 K) - 49.9 g × 1 cal / g°C × (19 K) ] / 19K
Ccal = 23.6 cal/ K
- Convert to cal / K to Joule / K
23.6 cal / K × 4.18 J / cal = 98.6 J/K
Which rounded to 2 signficant figures leads to 99 J/k, which is the first choice.