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oee [108]
3 years ago
6

How much of a 0.74mg sample of uranium-235 will remain after 2.8x10^9

Chemistry
1 answer:
ankoles [38]3 years ago
5 0
Of Uranium-235 remains after 2.8 x 10^9 years, what was the original mass of the sample of Uranium-235? The half-life of Uranium-235 is 7.0 x 10^8 years. Uranium-232 has a half life of 68.8 years.
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Calculate the molar concentration of the Br⁻ ions in 0.51 M MgBr2(aq), assuming that the dissolved substance dissociates complet
Y_Kistochka [10]
MgBr2(aq) is an ionic compound which will have the releasing of 2 Br⁻ ions ions in water for every molecule of MgBr2 that dissolves.
MgBr2(s) --> Mg+(aq) + 2 Br⁻(aq)
            [Br⁻] = 0.51 mol MgBr2/1L × 2 mol Br⁻ / 1 mol MgBr2 = 1.0 M
The answer to this question is [Br⁻] = 1.0 M
6 0
3 years ago
The organ system that allows for movement of the organism as a whole an for internal movement of food through the digestive syst
Y_Kistochka [10]

Answer:

muscular system

Explanation:

8 0
3 years ago
When 32 grams of aluminum react, the actual yield is 105.5 grams, what is the percent yield?
user100 [1]

Answer:

329.7%

Explanation:

Percent Yield = Actual Yield/ Theoretical Yield x 100%

Percent Yield = 105.5g/32 x 100% = 329.69 ≈ 329.7 %

5 0
3 years ago
How does the VSEPR theory explain molecular shape?
Advocard [28]

Answer:

D. Electron pairs repelling each other push atoms apart

Explanation:

Hope this helps :) I just got it wrong on ap3x so I'm sure this is right

3 0
3 years ago
Read 2 more answers
Calculate the mass in micrograms of zinc oxalate the chemist has added to the flask. be sure your answer has the correct number
Vera_Pavlovna [14]
<span>A chemist adds 155.0ml of a 4.10 X 10^-5 mmol/L of a zinc oxalate (ZnC2O4)solution to a reaction flask. Calculate the mass in micrograms of zinc oxalate the chemist has added to the flask. 1mmol = 10^-3 mol Therefore 4.10*10^-5mmol = 4.10*10^-8mol molar mass ZnC2O4 = 65.39+(2*12.011)+(4*15.99) = 153.372g/mol You have 4.10*10^-8 mol/litre =153.372 * 4.10*10^-8 = 6.29*10^-6 grams / litre (* see below) But you have 155ml. Mass of ZnC2O4 = 155/1000*6.29*10^-6 g Mass is = 9.75*10^-7 grams 1µg = 10^-6 g You then have 9.75*10^-7/10^-6 = 0.975µg ZnC2O4 (*see below) at this point you could have said: 1µg = 10^-6 g therefore you have a solution of 6.29µg per litre, 155ml = 6.29*155/1000 = 0.975µg ZnC2O4</span>
3 0
3 years ago
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