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Dvinal [7]
2 years ago
13

The table shows the amount of radioactive element remaining in a sample over a period of time.

Chemistry
1 answer:
Assoli18 [71]2 years ago
7 0

It would take 147 hours for 320 g of the sample to decay to 2.5 grams from the information provided.

Radioactivity refers to the decay of a nucleus leading to the spontaneous emission of radiation. The half life of a radioactive nucleus refers to the time required for the nucleus to decay to half of its initial amount.

Looking at the table, we can see that the initial mass of radioactive material present is 186 grams, within 21 hours, the radioactive substance decayed to half of its initial mass (93 g). Hence, the half life is 21 hours.

Using the formula;

k = 0.693/t1/2

k = 0.693/21 hours = 0.033 hr-1

Using;

N=Noe^-kt

N = mass of radioactive sample at time t

No = mass of radioactive sample initially present

k = decay constant

t = time taken

Substituting values;

2.5/320= e^- 0.033 t

0.0078 = e^- 0.033 t

ln (0.0078) = 0.033 t

t = ln (0.0078)/-0.033

t = 147 hours

Learn more: brainly.com/question/6111443

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5 0
3 years ago
How much 10.0 M HNO must be added to 1.00 L of a buffer that is 0.0100 M acetic acid and 0.100 M sodium acetate to reduce the pH
Naya [18.7K]

Explanation:

It is given that molarity of acetic acid = 0.0100 M

Therefore, moles of acetic acid = molarity of acetic acid × volume of buffer

            Moles of acetic acid = 0.0100 M × 1.00 L

                                              = 0.0100 mol

Similarly, moles of acetate = molarity of sodium acetat × volume of buffer

                                           = 0.100 mol

When HNO_{3} is added, it will convert acetate to acetic acid.

Hence, new moles acetic acid = (initial moles acetic acid) + (moles HNO_{3})

                                                = 0.0100 mol + x

New moles of sodium acetate = (initial moles acetate) - (moles HNO_{3})

                                        = 0.100 mol - x

According to Henderson - Hasselbalch equation,

           pH = pK_{a} + log\frac{[conjugate base]}{[weak acid]}

             pH = pKa + log\frac{(new moles of sodium acetate)}{(new moles of acetic acid)}

           4.95 = 4.75 + log\frac{(0.100 mol - x)}{(0.0100 mol + x)}

       log\frac{(0.100 mol - x)}{(0.0100 mol + x)}  = 4.95 - 4.75

                                            = 0.20

\frac{(0.100 mol - x)}{(0.0100 mol + x)}  = antilog (0.20)

                                           = 1.6

Hence,    x = 0.032555 mol

Therefore, moles of HNO_{3} = 0.032555 mol

volume of HNO_{3} = \frac{moles HNO_{3}}{molarity of HNO_{3}}

                                = \frac{0.032555 mol}{10.0 M}

                                 = 0.0032555 L

or,                             = 3.25           (as 1 L = 1000 mL)

Thus, we can conclude that volume of HNO_{3} added is 3.26 mL.

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