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zloy xaker [14]

2 years ago

14

What is the proper way of collection, preservation and transition of hair?

1 answer:

Murrr4er [49]2 years ago

7 0

An investigator can collect hairs they observe visually (with tweezers or by hand), and they can also use clear tape to lift non-visible hair from a variety of surfaces, such as clothing. Other methods of hair sample collection include combing and clipping methods.

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A sample contains 25% parent isotope and 75% daughter isotopes. If the half-life of the parent isotope is 72 years, how old is t

alisha [4.7K]

The radioactive decay obeys first order kinetics

the rate law expression for radioactive decay is

$ln\frac{[A_{0}]}{[A_{t}]}=kt$

Where

A0 = initial concentration

At = concentration after time "t"

t = time

k = rate constant

For first order reaction the relation between rate constant and half life is:

$k=\frac{0.693}{t_{\frac{1}{2} } }$

Let us calculate k

k = 0.693 / 72 = 0.009625 years⁻¹

Given

At = 0.25 A0

$ln(\frac{A0}{0.25A0})=0.009625 X time$

time = 144 years

So after 144 years the sample contains 25% parent isotope and 75% daughter isotopes**

Simply two half lives

5 0

2 years ago

Read 2 more answers

In an experiment you are asked to react hydrochloric acid with sodium hydroxide. When measuring the volume of the reactants, whi

Bingel [31]

What are the options? based on the volume it can differ but in general pipettes and graduated cylinder are the best. <span />

7 0

3 years ago

A sample of O2 gas occupies a volume of 571 mL at 26 ºC. If pressure remains constant, what would be the new volume if the tempe

Vlad1618 [11]

Answer: The new volume at different given temperatures are as follows.

(a) 109.81 mL

(b) 768.65 mL

(c) 18052.38 mL

Explanation:

Given: $V_{1}$ = 571 mL, $T_{1} = 26^{o}C$

(a) $T_{2} = 5^{o}C$

The new volume is calculated as follows.

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{5^{o}C}\\V_{2} = 109.81 mL$

(b) $T_{2} = 95^{o}F$

Convert degree Fahrenheit into degree Cesius as follows.

$(1^{o}F - 32) \times \frac{5}{9} = ^{o}C\\(95^{o}F - 32) \times \frac{5}{9} = 35^{o}C$

The new volume is calculated as follows.

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{35^{o}C}\\V_{2} = 768.65 mL$

(c) $T_{2} = 1095 K = (1095 - 273)^{o}C = 822^{o}C$

The new volume is calculated as follows.

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\\frac{571 mL}{26^{o}C} = \frac{V_{2}}{822^{o}C}\\V_{2} = 18052.38 mL$

8 0

2 years ago

.00 mol of gas occupy a volume of 112 L at a pressure of 1.00 atm. What is the kelvin temperature of the gas

artcher [175]

Sorry i couldnt help

5 0

2 years ago

Can someone help me please?

morpeh [17]

This rock is balanced by roots on the ground that are very strong

8 0

2 years ago