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tankabanditka [31]
3 years ago
15

If there is currently 50kg of U-235 present in Oklo, how much must have been present 750 million years ago when the reaction too

k place?
Chemistry
1 answer:
Viefleur [7K]3 years ago
7 0
To answer this question, you need to know the concept of half-life, which is how a radioactive material decreases in mass over time.

The half life of U-235 is 703.8 million years. The first part of this problem is to find the scale factor. To do this, divide the time that has past by the half life, like this:

\frac{750}{703.8}  = 1.066
Now, take this scale factor and multiply it by the current mass, like this:

50 \times 1.066 = 53.3
This number is what you add to the current mass to get the original mass. That is because the scale factor showed us that it was just over one half life. Since after one half life, the mass is cut in half, and this is over one half life, when we add to the original it will be a little over double. This equation illustrates the final addition:

50 + 53.3 = 103.3
I hope this helped you. Fell free to ask any further questions.
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Describe how the particles of two substances interact when mixed. Be sure to discuss the activation energy and how the particles
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It is common knowledge that chemical reactions occur more rapidly at higher temperatures. Everyone knows that milk turns sour much more rapidly if stored at room temperature rather than in a refrigerator, butter goes rancid more quickly in the summer than in the winter, and eggs hard-boil more quickly at sea level than in the mountains. For the same reason, cold-blooded animals such as reptiles and insects tend to be noticeably more lethargic on cold days.

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8 0
3 years ago
A 25.0 ml sample of 0.150 m hydrazoic acid is titrated with a 0.150 m naoh solution. what is the ph after 15.0 ml of the sodium
babunello [35]
First, we need to calculate moles of hydrazoic acid NH3:

moles NH3 = molarity * volume 

                    = 0.15 m * 0.025 L

                   =  0.00375 moles

moles NaOH = molarity * volume 

                       = 0.15 m * 0.015 L

                       = 0.00225 moles 

after that we shoul get the total volume = 0.025L + 0.015L

                                                                   = 0.04 L

So we can get the concentration of NH3 & NaOH by:

∴[NH3] = moles NH3 / total volume 

           = 0.00375 moles / 0.04 L

           = 0.09375 M

∴[NaOH] = moles NaOH / total volume 

                = 0.00225 moles / 0.04 L

                = 0.05625 M

then, when we have the value of Ka of NH3 so we can get the Pka value from:

Pka = -㏒Ka 

       = - ㏒ 1.9 x10^-5

      = 4.7 

finally, by using H-H equation we can get PH:

PH = Pka + ㏒[salt/ basic]

PH = 4.7 +㏒[0.05625/0.09375]

∴ PH = 4.48 


6 0
3 years ago
How to balance Kmno4 + H2 So4+ CH3CH2OH-> K2So4+ mnso4 + CH3COOH + H2o by oxidation number method ​
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Answer:

Ethanol + Potassium Permanganate + Sulfuric Acid = Acetic Acid + Manganese(II) Sulfate + Water + Potassium Sulfate

Explanation:

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3 years ago
How many moles are in 3.113 g of Au?Molar mass of Au=197 g/mol
SVEN [57.7K]
<h3>Answer:</h3>

0.0157 g Au

<h3>General Formulas and Concepts:</h3>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right
<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.113 g Au

<u>Step 2: Identify Conversions</u>

Molar Mass of Au - 197.87 g/mol

<u>Step 3: Convert</u>

<u />3.113 \ g \ Au(\frac{1 \ mol \ Au}{197.87 \ g \ Au} ) = 0.015733 g Au

<u>Step 4: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules and round.</em>

0.015733 g Au ≈ 0.0157 g Au

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Any type of medical scientist works.
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