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4 years ago

When is the mathematical label used?

Chemistry

1 answer:

Tamiku [17]4 years ago

7 0

Answer:

The mathematical label is used to represent angles of equal measure.

Explanation:

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Which statement best describes a typical difference that could be found between the "Analysis" and "Conclusion" sections of a

IgorC [24]

The statement “Only the “Conclusion” section discusses whether the original hypothesis was supported, and both sections suggest further research”, best describes the difference between analysis and conclusion.

Answer: Option 4

<u>Explanation: </u>

In research, we do experiments and derive the results. Then, those results were analyzed by us. In this analysis part, we compare our results with the related results published elsewhere. Also, we correlate the similarities and point out the differences between our analysis and other reported results.

In conclusion part, we have to check hypothesis or it supported. And, we summarise our analysis and figure out the further research need to be done on that to improvise our research. So, the final statement is the correct option which best describes the difference between analysis and conclusion.

3 0

3 years ago

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

_____ are the results of a thoroughly tested hypothesis?

Andru [333]

i think its theory but im not to sure

6 0

3 years ago

The threat of another hurricane is a problem for New York City. What have people

Maksim231197 [3]

Answer:

idk

Explanation:

4 0

3 years ago

Read 2 more answers

Calculate the heat change involved when 2.00 L of water is heated from 20.0/C to 99.7/C in

IgorC [24]

Answer:

669.48 kJ

Explanation:

According to the question, we are required to determine the heat change involved.

We know that, heat change is given by the formula;

Heat change = Mass × change in temperature × Specific heat

In this case;

Change in temperature = Final temp - initial temp

= 99.7°C - 20°C

= 79.7° C

Mass of water is 2000 g ( 2000 mL × 1 g/mL)

Specific heat of water is 4.2 J/g°C

Therefore;

Heat change = 2000 g × 79.7 °C × 4.2 J/g°C

= 669,480 joules

But, 1 kJ = 1000 J

Therefore, heat change is 669.48 kJ

7 0

3 years ago