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lawyer [7]
3 years ago
9

The half life for the decay of carbon- is years. Suppose the activity due to the radioactive decay of the carbon- in a tiny samp

le of an artifact made of wood from an archeological dig is measured to be . The activity in a similar-sized sample of fresh wood is measured to be . Calculate the age of the artifact. Round your answer to significant digits.

Chemistry
1 answer:
EleoNora [17]3 years ago
6 0

The question is incomplete. The complete question is:

The half-life for the decay of carbon-14 is 5.73x10^3 years. Suppose the activity due to the radioactive decay of the carbon-14 in a tiny sample of an artifact made of woodfrom an archeological dig is measured to be 2.8x10^3 Bq. The activity in a similiar-sized sample of fresh wood is measured to be 3.0x10^3 Bq. Calculate the age of the artifact. Round your answer to 2 significant digits.

Answer:

570 years

Explanation:

The activity of the fresh sample is taken as the initial activity of the wood sample while the activity measured at a time t is the present activity of the wood artifact. The time taken for the wood to attain its current activity can be calculated from the formula shown in the image attached. The activity at a time t must always be less than the activity of a fresh wood sample. Detailed solution is found in the image attached.

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Answer:

H^++NH_3\rightleftharpoons NH_4^+

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H^++NH_3\rightleftharpoons NH_4^+

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One type of bacteria reproduces once every 60 minutes. If there are 2 bacterial cells to begin with, then after 4 hours there wi
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Solving by the method of exponential growth.
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Find the pHpH of a solution prepared from 1.0 LL of a 0.15 MM solution of Ba(OH)2Ba(OH)2 and excess Zn(OH)2(s)Zn(OH)2(s). The Ks
charle [14.2K]

Answer:

pH  = 13.09

Explanation:

Zn(OH)2 --> Zn+2 + 2OH-   Ksp = 3X10^-15

Zn+2 + 4OH-   --> Zn(OH)4-2   Kf = 2X10^15

K = Ksp X Kf

  = 3*2*10^-15 * 10^15

  = 6

Concentration of OH⁻ = 2[Ba(OH)₂] = 2 * 0.15 = 3 M

                Zn(OH)₂ + 2OH⁻(aq)  --> Zn(OH)₄²⁻(aq)

Initial:           0             0.3                      0

Change:                      -2x                     +x

Equilibrium:               0.3 - 2x                 x

K = Zn(OH)₄²⁻/[OH⁻]²

6 = x/(0.3 - 2x)²  

6 = x/(0.3 -2x)(0.3 -2x)

6(0.09 -1.2x + 4x²) = x

0.54 - 7.2x + 24x² = x

24x² - 8.2x + 0.54 = 0

Upon solving as quadratic equation, we obtain;

x = 0.089

Therefore,

Concentration of (OH⁻) = 0.3 - 2x

                                    = 0.3 -(2*0.089)

                                  = 0.122

pOH = -log[OH⁻]

         = -log 0.122

          = 0.91

pH = 14-0.91

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4 0
3 years ago
PLS HELP: CHEMISTRY
user100 [1]

Answer:

3.59x10^21 molecules

Explanation:

1mole of a substance contains 6.02x10^23 molecules.

Therefore, 1mole of C8H18 will also contain 6.02x10^23 molecules

1mole of C8H18 = (12x8) +(18x1) = 96 + 18 = 114g.

1mole (i.e 114g) oh C8H18 contains 6.02x10^23 molecules.

Therefore, 0.68g of C8H18 will contain = (0.68 x 6.02x10^23)/114 = 3.59x10^21 molecules

7 0
3 years ago
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