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Hitman42 [59]
3 years ago
5

A scientist wrote a report that communicates the steps he performed in an investigation, the data he gathered, and his conclusio

ns. A second scientist used the report to carry out the same investigation. However, her results were very different. Are the results of the investigation described below valid? If not, what should the scientists do next? Explain your reasoning.
Chemistry
1 answer:
Nookie1986 [14]3 years ago
4 0

Answer:

<u>yes, this is an acceptable practice in research.</u>

<u>Explanation:</u>

Note that, having a different result in one's research does not imply it is automatically invalid. However, because there may have been some errors or limitations in the previous work by the male researcher, it is possible that the later results are more accurate, even though she used the same procedures. She may have been more careful when calculating her results.

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What product is obtaned when carbon is
kirza4 [7]

Answer:

carbon dioxide

Explanation:

Carbon burns in oxygen to form carbon dioxide. Since hydrocarbon fuels only contain two elements, we always obtain the same two products when they burn. In the equation below methane (CH 4) is being burned. The oxygen will combine with the carbon and the hydrogen in the methane molecule to produce carbon dioxide (CO 2) and water (H 2O).

Carbon, as graphite, burns to form gaseous carbon (IV) oxide (carbon dioxide), CO2. ... When the air or oxygen supply is restricted, incomplete combustion to carbon monoxide, CO, occurs. 2C(s) + O2(g) → 2CO(g) This reaction is important. When one mole of carbon is exposed to some energy in the presence of one mole of oxygen gas, one mole of carbon dioxide gas is produced. This reaction is a combustion reaction.

6 0
3 years ago
An alloy with an average grain diameter of 35 μm has a yield strength of 163 Mpa, and when it undergoes strain hardening, the gr
Lera25 [3.4K]

Answer:

\sigma_y\ =210.2\ MPa  

Explanation:

Given that

d= 35 μm ,yield strength = 163 MPa

d= 17 μm ,yield strength = 192 MPa

As we know that relationship between diameter and yield strength

\sigma_y=\sigma_o+\dfrac{K}{\sqrt d}

\sigma_y\ =Yield\ strength

d = diameter

K =Constant

\sigma_o\ =material\ constant

So now by putting the values

d= 35 μm ,yield strength = 163 MPa

163=\sigma_o+\dfrac{K}{\sqrt 35}      ------------1

d= 17 μm ,yield strength = 192 MPa

192=\sigma_o+\dfrac{K}{\sqrt 17}           ------------2

From equation 1 and 2

192-163=\dfrac{K}{\sqrt 17}-\dfrac{K}{\sqrt 35}

K=394.53

By putting the values of K in equation 1

163=\sigma_o+\dfrac{394.53}{\sqrt 35}

\sigma_o\ =96.31\ MPa

\sigma_y=96.31+\dfrac{394.53}{\sqrt d}

Now when d= 12 μm

\sigma_y=96.31+\dfrac{394.53}{\sqrt 12}

\sigma_y\ =210.2\ MPa

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