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4 years ago

10

You have decided to do a science fair project involving plants. Which of the following represents a well-stated hypothesis?

2 answers:

marysya [2.9K]4 years ago

8 0

The correct answer should be c

Brums [2.3K]4 years ago

3 0

the correct answer would definitly be c , hope this helps

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Which part of an investigation is only found in an experimental investigation?

pantera1 [17]

Answer:

The correct answer is control group.

Explanation:

A group used in a study or in an experiment, which does not get treatment by the scientists and is used as a foundation to determine the functions of the other tested subjects is known as the control group. The control group is only found in an experimental investigation.

The group in an experiment, which gets the variable being examined is known as an experimental group. The comparison of an experimental group is done with a control group in order to find the answers in an experiment.

3 0

3 years ago

a reaction in which one element replaces another on a compound or in which two elements in different compounds trade places is c

lilavasa [31]

Single Replacement and Double Replacement reactions

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4 years ago

What improvement would you make?

galben [10]

I think is it the flask the bottle cause i had the same thing and i got it right idk maybe :)

6 0

3 years ago

A galvanic cell at a temperature of 42 degrees Celcius is powered by the following redox reaction:

seropon [69]

<u>Answer:</u> The cell voltage of the given reaction is 1.86 V

<u>Explanation:</u>

The given chemical equation follows:

$3Cu^{2+}(aq.)+2Al\rightarrow 2Al^{3+}(aq.)+2Au(s)$

<u>Oxidation half reaction:</u> $Al(aq.)\rightarrow Al^{3+}(aq.)+3e^-;E^o_{Al^{3+}/Al}=1.66V$ ( × 2)

<u>Reduction half reaction:</u> $Cu^{2+}(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.16V$ ( × 3)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

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

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

Putting values in above equation, we get:

$E^o_{cell}=0.16-(-1.66)=1.82V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Al^{3+}]^2}{[Cu^{2+}]^3}$

where,

$E_{cell}$ = electrode potential of the cell = ? V

$E^o_{cell}$ = standard electrode potential of the cell = +1.82 V

n = number of electrons exchanged = 2

R = Gas constant = 8.314 J/mol Kl

T = temperature = $42^oC=[42+273]K=315K$

F = Faraday's constant = 96500

$[Al^{3+}]=1.63M$

$[Cu^{2+}]=3.43M$

Putting values in above equation, we get:

$E_{cell}=1.82-\frac{2.303\times 8.314\times 315}{2\times 96500}\times \log(\frac{(1.63)^2}{(3.43)^3})\\\\E_{cell}=1.86V$

Hence, the cell voltage of the given reaction is 1.86 V

4 0

3 years ago

How many moles of iron in a sample that contains 7.91×10^23 atoms of iron?

vazorg [7]

Answer: 1.31

Explanation:

No.of moles = given no.of atoms/Avagadro number

= 7.91×10^23 / 6.022 x 10^23

= 1.31

therefore, no.of moles = 1.31

Hope it helped u,

pls mark as the brainliest

^_^

5 0

3 years ago