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mafiozo [28]
3 years ago
8

The half life of carbon 14 is 5730 years what fraction of the original c14 would you expect to be present in a fossil that is 28

,650 years old
Chemistry
1 answer:
Bogdan [553]3 years ago
7 0
The half-life of a radioisotope describes the amount of time required for one-half of the isotope to decay into a daughter atom. After one half-life, 50% remains; after two, 25%, and so on.

To find how many half-lives have passed in 28,650 years, we can divide it by the time it takes for one half-life to pass. This reveals that five half-lives have passed.

HL ———— 1 2 3 4 5
% remain - 50 25 12.5 6.25 3.125

After 28,650 years or 5 half-lives, 3.125% of the original radioisotope remains in the fossil.
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Explain why, when performing stoichiometric calculations, it is
pshichka [43]

Answer:

Stoichiometric Coefficients

The balanced equation makes it possible to convert information about one reactant or product to quantitative data about another element. Understanding this is essential to solving stoichiometric problems

Explanation:

3 0
3 years ago
The effectiveness of nitrogen fertilizers depends on both their ability to deliver nitrogen to plants and the amount of nitrogen
densk [106]

Answer:

Ammonia > Urea > Ammonium nitrate > Ammonium sulphate

Explanation:

Percentage by mass of nitrogen in NH3:

Molar mass of NH3= 17 g/mol

Hence % by mass = 14/17 × 100 = 82.35%

% by mass of NH4NO3

Molar mass of NH4NO3 = 80.043 g/mol

Hence; 28/80.043 × 100 = 34.98%

% by mass of (NH4)2SO4;

Molar mass of (NH4)2SO4= 132.14 g/mol

Hence; 28/132.14 × 100 = 21.19%

% by mass of CH4N2O

Molar mass of urea = 60.0553 g/mol

Hence 28/60.0553 × 100 = 46.62%

8 0
3 years ago
How many moles of iron is 6.022 x 10^22 atoms of iron? (Report answer as a number rounded to one place past the decimal.) *
yuradex [85]
<h3>Answer:</h3>

\displaystyle 0.1 \ mol \ Fe

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

<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

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

6.022 × 10²² atoms Fe (iron)

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

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                    \displaystyle 6.022 \cdot 10^{22} \ atoms \ Fe(\frac{1 \ mol \ Fe}{6.022 \cdot 10^{23} \ atoms \ Fe})
  2. Divide:                    \displaystyle 0.1 \ mol \ Fe
7 0
3 years ago
A bucket of water contains 5.45 L of water, how many mL of water is this ?
nalin [4]
Answer: It is 5450 mL


Explanation: There are 1000 mL in every L and then there is an extra 450 so just add that at the end
6 0
3 years ago
Sodium tert-butoxide (NaOC(CH3)3) is classified as bulky and acts as Bronsted Lowry base in the reaction. It is reacted with 2-c
IceJOKER [234]

Answer:but-1-ene

Explanation:This is an E2 elimination reaction .

Kindly refer the attachment for complete reaction and products.

Sodium tert-butoxide is a bulky base and hence cannot approach the substrate 2-chlorobutane from the more substituted end and hence major product formed here would not be following zaitsev rule of elimination reaction.

Sodium tert-butoxide would approach from the less hindered side that is through the primary centre and hence would lead to the formation of 1-butene .The major product formed in this reaction would be 1-butene .

As the mechanism of the reaction is E-2 so it will be a concerted mechanism and as sodium tert-butoxide will start abstracting the primary hydrogen through the less hindered side simultaneously chlorine will start leaving. As the steric repulsion in this case is less hence the transition state is relatively stabilised and leads to the formation of a kinetic product 1-butene.

Kinetic product are formed when reactions are dependent upon rate and not on thermodynamical stability.

2-butene is more thermodynamically6 stable as compared to 1-butene  

The major product formed does not follow the zaitsev rule of forming a more substituted alkene as sodium tert-butoxide cannot approach to abstract the secondary proton due to steric hindrance.

5 0
3 years ago
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