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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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We say that the solution is unsaturated.

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2. 2AgNO3 + CaCl2 → 2AgCl + Ca(NO3)2
Slav-nsk [51]

Answer:

84.4g of AgCl

Explanation:

Based on the reaction:

2AgNO₃ + CaCl₂ → 2AgCl + Ca(NO₃)₂

<em>2 moles of AgNO₃ and 1 mole of CaCl₂ priduce 2 moles of AgCl and 1 mole of Ca(NO₃)₂</em>

<em />

100g of each reactant are:

AgNO₃: 100g × (1mol / 169.87g) = 0.589 moles

CaCl₂: 100g × (1mol / 110.98g) = 0.901 moles

For a complete reaction of 0.901 moles of CaCl₂ are necessaries 0.901×2 = <em>1.802 moles of AgNO₃. </em>As there are just 0.589moles, <em>AgNO₃ is limitng reactant</em>

<em></em>

0.589 moles of AgNO₃ produce:

0.589 moles × ( 2 moles AgCl / 2 moles AgNO₃) =

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0.589 moles of AgCl × (143.32g / mol) =<em> 84.4g of AgCl</em>

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4 years ago
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