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Brut [27]
2 years ago
15

The half-life of carbon-14 is 5,730 years. Dating organic material by looking for C-14 can't be accurately done after 50,000 yea

rs. Suppose a fossilized tree branch originally contained 4.30 grams of C-14. How much C-14 would be left after 50,000 years?
Chemistry
1 answer:
ki77a [65]2 years ago
7 0

Answer:

0.010g of C-14 would be later after 50,000 years

Explanation:

The kinetics of radioactive decay follows the equation:

Ln (N / N₀) = -kt

<em>Where N could be taken as mass after time t, </em>

<em>N₀ initial mass = 4.30g;</em>

k is rate constant = ln 2 / t(1/2)

<em>= ln 2 / 5730years = 1.2097x10⁻⁴ years ⁻¹</em>

<em />

Replacing:

Ln (N / 4.30g) = -1.2097x10⁻⁴ years ⁻¹ * 50000 years

N / 4.30 = 2.36x10⁻³

N =

<h3>0.010g of C-14 would be later after 50,000 years</h3>
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3 0
3 years ago
Conserving water can save money while protecting the environment. True or False
blagie [28]

Answer: The given statement is true.

Explanation:

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3 0
3 years ago
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2. The Density of Water is 1.0000g/cm3. Based on this property of water,
Vanyuwa [196]

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if the object sank then that object has a greater density then water. if the object floated then its density is lower then water.

Explanation:

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7 0
3 years ago
How much heat is required to raise the temperature of 225 grams of ice from -26.8 °C to steam at 133 °C ?
lara [203]
<h3>Answer:</h3>

150000 J

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

<u>Chemistry</u>

<u>Thermodynamics</u>

Specific Heat Formula: q = mcΔT

  • <em>q</em> is heat (in J)
  • <em>m</em> is mass (in g)
  • <em>c</em> is specific heat (in J/g °C)
  • ΔT is change in temperature (in °C or K)

<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>

<em>Identify variables</em>

[Given] <em>m</em> = 225 g

[Given] <em>c</em> = 4.184 J/g °C

[Given] ΔT = 133 °C - -26.8 °C = 159.8 °C

[Solve] <em>q</em>

<u>Step 2: Solve for </u><em><u>q</u></em>

  1. Substitute in variables [Specific Heat Formula]:                                          q = (225 g)(4.184 J/g °C)(159.8 °C)
  2. Multiply:                                                                                                           q = (941.4 J/°C)(159.8 °C)
  3. Multiply:                                                                                                           q = 150436 J

<u>Step 3: Check</u>

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

150436 J ≈ 150000 J

Topic: AP Chemistry

Unit: Thermodynamics

Book: Pearson AP Chemistry

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