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Andrej [43]
2 years ago
6

The half-life of radon gas is approximately four days. Four weeks after the introduction of radon into a sealed room, the fracti

on of the original amount remaining is closest to
Chemistry
1 answer:
maksim [4K]2 years ago
7 0

The fraction of the original amount remaining is closest to 1/128

<h3>Determination of the number of half-lives</h3>
  • Half-life (t½) = 4 days
  • Time (t) = 4 weeks = 4 × 7 = 28 days
  • Number of half-lives (n) =?

n = t / t½

n = 28 / 4

n = 7

<h3>How to determine the amount remaining </h3>
  • Original amount (N₀) = 100 g
  • Number of half-lives (n) = 7
  • Amount remaining (N)=?

N = N₀ / 2ⁿ

N = 100 / 2⁷

N = 0.78125 g

<h3>How to determine the fraction remaining </h3>
  • Original amount (N₀) = 100 g
  • Amount remaining (N)= 0.78125 g
  • Fraction remaining =?

Fraction remaining = N / N₀

Fraction remaining = 0.78125 / 100

Fraction remaining = 1/128

Learn more about half life:

brainly.com/question/26374513

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How can I balance this equation? TELL ME HOW you did it and all the steps and you will be the brainliest.
lianna [129]

Answer:

3 CH3CH2OH + 4 H2CrO4 + 6 H2SO4 --> 3 CH3COOH + 2 Cr2(SO4)3 + 13 H2O

Explanation:

To balance, start of with the groups that are common on both sides of the reaction equation;

In this case, these are the SO4 groups treating them as single units;

There are 3 on the right side and 1 on the left side, so we put 3 in front of the H2SO4 to balance this first;

Next, deal with the Cr, there are 2 Cr on the right side and 1 on the other side, so we put a 2 in front of the H2CrO4 to balance that;

Thirdly, we notice the C are already balanced as there are 2 on each side so this is fine;

Lastly, we can deal with the O and H;

Bearing in mind the numbers that are in front of the molecules now from prior balancing, there are 9 O and 16 H on the left side and 3 O and 5 H on the right;

7 H2O on the right side would balance the O, but gives us 18 H, which is 2 too many H;

If we were to put 2 in front of the two organic molecules (the ones with C) on either side, we would balance the O by having 6 H2O, but this gives 2 fewer H than necessary;

In order for the H to balance, we need to have 13/2 (or 6.5) H2O, which means we need 3/2 (or 1.5) in front of each organic molecule;

Since, it is not sensible to have 13/2 water molecules or 3/2 organic molecules, we can just multiply everything by 2;

Thus we end up with:

3 CH3CH2OH + 4 H2CrO4 + 6 H2SO4 --> 3 CH3COOH + 2 Cr2(SO4)3 + 13 H2O

Rules of thumb:

- When there are common chemical groups (e.g. SO4) on both sides of a reaction equation, treat them as single units

- Start of with balancing these common groups

- Thereafter, balance the atoms that appear in only one reactant and one product

- Proceed to balance the atoms that appear in more than one reactant or product

- Typically, you should deal with the O and H last

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Answer:

⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣧⣀⣀⣾⣿⣿⣿⣿

⣿⣿⣿⣿⣿⡏⠉⠛⢿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⡿⣿

⣿⣿⣿⣿⣿⣿⠀⠀⠀⠈⠛⢿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⣿⠿⠛⠉⠁⠀⣿

⣿⣿⣿⣿⣿⣿⣧⡀⠀⠀⠀⠀⠙⠿⠿⠿⠻⠿⠿⠟⠿⠛⠉⠀⠀⠀⠀⠀⣸⣿

⣿⣿⣿⣿⣿⣿⣿⣷⣄⠀⡀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢀⣴⣿⣿

⣿⣿⣿⣿⣿⣿⣿⣿⣿⠏⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠠⣴⣿⣿⣿⣿

⣿⣿⣿⣿⣿⣿⣿⣿⡟⠀⠀⢰⣹⡆⠀⠀⠀⠀⠀⠀⣭⣷⠀⠀⠀⠸⣿⣿⣿⣿

⣿⣿⣿⣿⣿⣿⣿⣿⠃⠀⠀⠈⠉⠀⠀⠤⠄⠀⠀⠀⠉⠁⠀⠀⠀⠀⢿⣿⣿⣿

⣿⣿⣿⣿⣿⣿⣿⣿⢾⣿⣷⠀⠀⠀⠀⡠⠤⢄⠀⠀⠀⠠⣿⣿⣷⠀⢸⣿⣿⣿

⣿⣿⣿⣿⣿⣿⣿⣿⡀⠉⠀⠀⠀⠀⠀⢄⠀⢀⠀⠀⠀⠀⠉⠉⠁⠀⠀⣿⣿⣿

⣿⣿⣿⣿⣿⣿⣿⣿⣧⠀⠀⠀⠀⠀⠀⠀⠈⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢹⣿⣿

⣿⣿⣿⣿⣿⣿⣿⣿⣿⠃⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⢸⣿⣿

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