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3 years ago

3. The scientific study of how organisms are classified is called _____.

1 answer:

3 0

Hey mate!

Taxonomy is the <span>study of classification of organisms. Therefore, your answer is A.

Hope this helps!</span>

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Since the half-life of 235U (7. 13 x 108 years) is less than that of 238U (4.51 x 109 years), the isotopic abundance of 235U has

Ymorist [56]

Answer:

$\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

Explanation:

Given that:

The Half-life of $^{235}U$ = $7.13 \times 10^8 \ years$ is less than that of $^{238} U = 4.51 \times 10^9 \ years$

Although we are not given any value about the present weight of $^{235}U$ .

So, consider the present weight in the percentage of $^{235}U$ to be y%

Then, the time elapsed to get the present weight of $^{235}U$ = $t_1$

Therefore;

$N_1 = N_o e^{-\lambda \ t_1}$

here;

$N_1$ = Number of radioactive atoms relating to the weight of y of $^{235}U$

Thus:

$In( \dfrac{N_1}{N_o}) = - \lambda t_1$

$In( \dfrac{N_o}{N_1}) = \lambda t_1$ --- (1)

However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of $^{235}U$ to be = $t_2$

Then:

$In( \dfrac{N_o}{N_2}) = \lambda t_2$ ---- (2)

here;

$N_2$ = Number of radioactive atoms of $^{235}U$ relating to 3.0 a/o weight

Now, equating equation (1) and (2) together, we have:

$In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) = \lambda( t_1-t_2)$

replacing the half-life of $^{235}U$ = $7.13 \times 10^8 \ years$

$In( \dfrac{N_2}{N_1}) = \dfrac{In 2}{7.13 \times 10^9}( t_1-t_2)$ ( since $\lambda = \dfrac{In 2}{t_{1/2}}$ )

∴

$\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}$

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o

Thus, The time elapsed is $\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}$

8 0

3 years ago

What is relative isotopic mass

lidiya [134]

Answer:

Relative atomic mass or atomic weight is a dimensionless physical quantity defined as the ratio of the average mass of atoms of a chemical element in a given sample to the atomic mass constant. The atomic mass constant is defined as being 1/12 of the mass of a carbon-12 atom.

8 0

3 years ago

How much heat is gained when a 50.32g piece of aluminum is heated from 9.0°c to 16°c

Rashid [163]

Answer: 317 joules

Explanation:

The quantity of heat energy (Q) gained by aluminium depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

In this case,

Q = ?

Mass of aluminium = 50.32g

C = 0.90J/g°C

Φ = (Final temperature - Initial temperature)

= 16°C - 9°C = 7°C

Then, Q = MCΦ

Q = 50.32g x 0.90J/g°C x 7°C

Q = 317 joules

Thus, 317 joules of heat is gained.

5 0

3 years ago

Read 2 more answers

Need help on question 3 plz<br> URGENT HELP PLZ

Law Incorporation [45]

Alloys are the homogeneous mixture of metals and non metals.... they are used in making cars and jewellery to make them more featurable,,, means that they would possess high regidity to the destruction .... both features of metal and nonmetal

5 0

3 years ago

Spell out the full name of the compound.

Talja [164]

Answer:

2–methyl–3–propylheptane.

Explanation:

The name of the compound can be obtained as follow:

1. Identify the functional group of the compound.

2. Determine the name of the compound by determining the longest continuous carbon chain.

3. Identify the substituent group attached.

4. Locate the position of the substituent group by giving it the lowest possible count. Where there are two or more different substitutent group attached to the compound, name them alphabetically.

5. Name the compound by combining

the above steps.

Thus, we can name the compound as follow:

1. The compound has only single bond. Therefore, the compound is an alkane.

2. The longest continuous carbon chain is 7. Therefore, the parent name of the compound is heptane.

3. The substitutent groups attached are:

Propyl i.e –C3H7

Methyl i.e –CH3

4. The substituent groups are located at:

Propyl i.e –C3H7 is located at carbon 3.

Methyl i.e –CH3 is located at carbon 2.

5. The name of the compound is:

2–methyl–3–propylheptane.

6 0

3 years ago