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3 years ago

How was or wasn't the scientific method followed in this experment

Chemistry

1 answer:

Andreas93 [3]3 years ago

3 0

Answer:

The best answer would be:

C. The scientific method was followed because the experiment tested the hypothesis and produced reliable results.

Explanation:

The purpose of an experiment is to test the hypothesis. It is true that having an experimental group and a control group was following the scientific method but the data gathering and documentation was also part of the scientific method. So the best answer would be C.

Writing the question is not necessary. It is merely part of forming a hypothesis. The scientific method does not require many scientists to perform the experiment, but it does require that the experiment be repeatable, so it can be tested again.

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What does E=mc^2 mean

Brums [2.3K]

E represents units of energy, m represents units of mass, and c2 is the speed of light squared, or multiplied by itself

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3 years ago

What is the specific heat of a substance if 1450 calories are required to raise the temperature of a 240g sample by 20℃?

bazaltina [42]

Answer:

6960 J/kg°C

Explanation:

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2 years ago

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Natasha2012 [34]

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3 years ago

Which of these is most likely to form between elements transferring electrons to

Stella [2.4K]

Answer: The answer is an ionic bond

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3 years ago

Read 2 more answers

For the following electrochemical reaction: Al3+(aq) + 3e -> Al(s) Eº = -1.66 V E° = 2.87 F2(g) + 2e -> 2F (aq) Calculate

makvit [3.9K]

<u>Answer:</u> The standard electrode potential of the cell is 4.53 V.

<u>Explanation:</u>

We are given:

$E^o_{(F_2/F^-)}=2.87V\\E^o_{(Al^{3+}/Al)}=-1.66V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Aluminium will undergo oxidation reaction and will get oxidized.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=2.87-(-1.66)=4.53V$

Hence, the standard electrode potential of the cell is 4.53 V.

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2 years ago