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ki77a [65]
2 years ago
10

Two students conduct the same experiment, but they get different results. What should the students do next?A.The students should

check their procedure and tools for sources of error. B.Both students should throw out their data and start over. C.The students should compare data with other students and then use the data from the student who was closest to the rest of the class. D.The students should find the average of each data set and each report the average data.
Chemistry
2 answers:
blsea [12.9K]2 years ago
4 0

Answer:

A.The students should check their procedure and tools for sources of error.

Explanation:

alexandr1967 [171]2 years ago
3 0

Answer:

I think it A

Explanation:

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A substance where almost all have the same atomic number of protons
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Suppose you have an Internet friend who study chemistry just like you are your friend claims the change from liquid water to wat
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3 years ago
A sample of milk kept at 25 °C is found to sour 40 times as rapidly as when it is kept at 4 °C. Estimate the activation energy f
irga5000 [103]

Answer:

120.575 kJ is the activation energy for the souring process.

Explanation:

The formula for an activation energy is given as:

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_1 = rate constant at 25^oC = 40k

K_2 = rate constant at 4^oC = k

Ea = activation energy for the reaction = ?

R = gas constant = 8.314 J/mole.K

T_1 = initial temperature = 25^oC=273+25=298 K

T_2 = final temperature = 4^oC=273+4=277 K

Now put all the given values in this formula, we get:l

\log (\frac{k}{40k})=\frac{Ea}{2.303\times 8.314 J/mol K}[\frac{1}{298K}-\frac{1}{277 K}]

E_a=120,575.61J=120.575 kJ

120.575 kJ is the activation energy for the souring process.

7 0
3 years ago
A 1.00 cm2 blackbody surface emits 201 watts of radiant power. What is the temperature of the surface? Give to the nearest whole
erastovalidia [21]

Explanation:

Formula for black body radiation is as follows.

                 \frac{P}{A} = \sigma \times T^{4} J/m^{2}s

where,       P = power emitted

                 A = surface area of black body

          \sigma = Stephen's constant = 5.6703 \times 10^{-8} watt/m^{2}.K^{-4}

As area is given as 1.0 cm^{2}. Converting it into meters as follows.

           1.00 cm^{2} \times \frac{(10^{-2})^{2} m^{2}}{1 cm^{2}}      (as 1 m = 100 cm)

            = 1 \times 10^{-4} m^{2}

It is given that P = 201 watts. Hence,  

       \frac{201 watts}{1 \times 10^{-4} m^{2}} = 5.6703 \times 10^{-8} watt/m^{2}.K^{-4} \times T^{4}

                T^{4} = 35.45 \times 10^{12}

                       T = (35.45 \times 10^{12})^{1/4}

                          = 8862.5 K

Thus, we can conclude that the temperature of the surface is 8862.5 K.

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A shorter electromagnetic wave is _____.
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