Answer:
6.25 g
Explanation:
From the question given above, the following data were obtained:
Half-life (t½) = 68.8 years
Time (t) = 344 years
Original amount (N₀) = 200 g
Amount remaining (N) =?
Next, we shall determine the number of half-lives that has elapsed. This can be obtained as follow:
Half-life (t½) = 68.8 years
Time (t) = 344 years
Number of half-lives (n) =
n = t / t½
n = 344 / 68.8
n = 5
Thus, 5 half-lives has elapsed.
Finally, we shall determine the amount of the Uranium-232 that remains. This can be obtained as follow:
Original amount (N₀) = 200 g
Number of half-lives (n) = 5
Amount remaining (N) =?
N = 1/2ⁿ × N₀
N = 1/2⁵ × 200
N = 1/32 × 200
N = 200 / 32
N = 6.25 g
Thus, the amount of Uranium-232 that remains is 6.25 g
group 1 elements are metals with<u> low</u> density
Answer: 218.75 kPa
Explanation:
Carbon dioxide is a gas with chemical formula CO2.
Original volume of CO2 (V1) = 3.50 L
Original pressure of CO2 (P1) = 125 kPa
New pressure of CO2 (P1) = ?
New Volume of CO2 (V2) = 2.0 L
Since pressure and volume are given while temperature is constant, apply the formula for Boyle's law
P1V1 = P2V2
3.50L X 125 kPa = P1 x 2.0L
437.5 L•kPa = 2.0L•P1
Divide both sides by 2.0L
437.5 L•kPa/2.0L = 2.0L•P1/2.0L
218.75 kPa = P1
Thus, the new pressure of carbon dioxide would be 218.75 kPa
AlBr3(aq) is an ionic compound which will have the releasing of 3 Br⁻ ions ions in water for every molecule of AlBr3 that dissolves.
AlBr3(s) --> Al+(aq) + 3 Br⁻(aq)
[Br⁻] = 0.16 mol AlBr3/1L × 3 mol Br⁻ / 1 mol AlBr3 = 0.48 M
The answer to this question is [Br⁻] = 0.48 M
Answer:How many moles of NH3 are produced from 9 moles of hydrogen? a mot H, 12 mol NH3 = 16 mol NH₂). 3 motta. 3. How many moles of nitrogen are needed to make 6 moles of NH3? ... a. masses, in grams, of all reactants and products.
Explanation:
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