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Ray Of Light [21]
3 years ago
10

Thorium-234 has a half-life of 24.1 days. How much of a 100-g sample of thorium-234 will be unchanged after 48.2 days?

Chemistry
2 answers:
oee [108]3 years ago
7 0
Half-life means that the matter degrades by 50% at each interval. 48.2 days is 2 intervals.

So, it would go down by 50% in the first interval, making it 50g. Then it would go down by another 50% from there, making the final amount of undecayed thorium-234 25g.
musickatia [10]3 years ago
5 0

25G        is the answer for sure.

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C. chemical reactions

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3 years ago
31,32<br> show work plz and help asap
Yakvenalex [24]

Answer:      31 (a) 4           32 (a) 5

                        (b) 7                 (b) 3

                        (c) 5                 (c) 5

                        (d) 3                 (d) 2

<u>Significant Figures (sf) Rules:</u>

1) Trailing zeroes are considered significant ONLY If a decimal point is provided.

2) Zeroes between the decimal point and number are NOT significant <u>unless</u> they are between numbers.

3) All digits in scientific notation are counted (disregard the exponent).

31 (a) 508.0              <em>trailing zero so use rule 1 </em>  -->  sf = 4

   (b) 820,400.0      <em>trailing zero so use rule 1 </em>  -->  sf = 7

   (c) 1.0200 x 10⁵   <em>scientific notation so use rule 3</em> --> sf = 5

    (d) 807,000         <em>no trailing zeroes so use rule 1</em> --> sf = 3

32 (a) 0.049450     <em>zero between decimal point & #, use rule 2</em> --> sf = 5

    (b) 0.000482     <em>zero between decimal point & #, use rule 2</em> --> sf = 3

    (c) 3.1587 x 10 ⁻⁸  <em>scientific notation so use rule 3</em> --> sf = 5

    (d) 0.0084          <em>zero between decimal point & #, use rule 2</em> --> sf = 2

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3 years ago
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3 years ago
In a reversible chemical reaction at equilibrium, the concentration of X (the reactant) is 0.75 mol/L, and the concentration of
frosja888 [35]

Answer : The value of K_{eq} is, 3

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

The given equilibrium reaction is,

X\rightleftharpoons Y

The expression of K_{eq} will be,

K_{eq}=\frac{[Y]}{[X]}

Given:

[X] = 0.75 mol/L

[Y] = 2.25 mol/L

Now put all the given values in the above expression, we get:

K_{eq}=\frac{2.25}{0.75}

K_{eq}=3

Therefore, the value of K_{eq} is, 3

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3 years ago
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