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Veronika [31]
3 years ago
11

The half life for the radioactive decay of rubidium-87 to strontium-87 is 4.88 × 10 years. Suppose nuclear chemical analysis sho

ws that there is 0.728 mmol of strontium-87 for every 1.000 mmol of rubidium-87 in a certain sample of rock. Calculate the age of the rock. Round your answer to 2 significant digits.
Chemistry
1 answer:
fredd [130]3 years ago
5 0

Answer:

Most igneous, metamorphic, and sedimentary rocks contain rubidium (Rb) and strontium (Sr) in detectable amounts. However, the concentrations of these elements are almost always less than 1 percent, and they are therefore rarely determined in routine chemical analyses. Neither rubidium nor strontium is a major constituent in the common rock-forming silicate minerals, although strontium does form a carbonate (strontianite) and a sulfate (celestite) which are found in some hydrothermal deposits and certain sedimentary rocks, particularly carbonates.

Explanation:

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The extent of sewage treatment required mostly depends on the control of suspended solids and BOD of the sewage. Hence, option C is correct.

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Hence, option C is correct.

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Explanation:

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Define the word isotope​
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What's the empirical and molecular weight of C14H22N4O8? What's the molecular formula of C14H22N4O8?
Inessa05 [86]

374u

187u

C₁₄H₂₂N₄O₈

Explanation:

To find the molecular weight of the compound C₁₄H₂₂N₄O₈ we simply sum that atomic masses of the given elements in the compound.

 The empirical weight is determined by using the simplest ratio of the elements involved in the compound;

Molecular weight of C₁₄H₂₂N₄O₈;

atomic mass of C = 12g/mol

                           H = 1g/mol

                            N = 14g/mol

                            O = 16g/mol

 Molecular weight = 14(12) + 22(1) + 4(14) + 8(16)

                                = 168 + 22 + 56 + 128

                                 = 374u

Empirical weight:

  Empirical formula:

                        C₁₄    H₂₂      N₄     O₈

                         14  :    22 :    4  :     8

   divide by 2:

                          7   :    11    :    2  :    4

     empirical formula  C₇H₁₁N₂O₄

     empirical weight = \frac{molecular weight}{2} = \frac{374}{2} = 187u

The molecular formula is the actual combination of atoms in a compound. so the molecular formula of the compound is C₁₄H₂₂N₄O₈

learn more:

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