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Talja [164]
3 years ago
8

How is a battery and a banana the same when considering energy?

Chemistry
1 answer:
emmainna [20.7K]3 years ago
4 0

A battery and a banana are the same when considering energy:

A battery when connected to a device produces electro-chemical reactions and contains one or more cells. The same thing can be tried out with fruits also. There are many fruits which can be taken into account. A lemon shows less amount of voltage when seen through a multimeter rather than a banana which shows a higher voltage. This was because of the acid present in the a fully ripped banana. The acid is called ascorbic acid. This acid has a higher pH level than compared to the pH of the citric acid found in lemon.  

This shows that a banana plant is not so expensive and reliable for the production of renewable energy and a battery is also inexpensive, reliable and stable.

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Why would scientists use a spectrometer in space.
solniwko [45]

Answer:

A. To study the chemical composition of the surface of a planet

Explanation:

Strictly speaking, a spectrometer is any instrument used to view and analyze a range (or a spectrum) of a given characteristic for a substance (for example, a range of mass-to-charge values as in mass spectrometry), or a range of wavelengths as in absorption spectrometry like nuclear magnetic radiation spectroscopy

-hope this helps!

6 0
2 years ago
Read 2 more answers
When 32 grams of aluminum react, the actual yield is 105.5 grams, what is the percent yield?
user100 [1]

Answer:

329.7%

Explanation:

Percent Yield = Actual Yield/ Theoretical Yield x 100%

Percent Yield = 105.5g/32 x 100% = 329.69 ≈ 329.7 %

5 0
3 years ago
¿Que atomo gana el electron?
antoniya [11.8K]

Answer:

anion

Explanation:

3 0
3 years ago
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kari74 [83]

Answer:

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6 0
3 years ago
4. A student started with a 0.032 g sample of copper which he took through the series of reactions described in this experiment.
Elodia [21]

Answer:

Y=48.6\%

Explanation:

Hello,

In this case, we can consider the following chemical reaction for the oxidation of copper which only occurs at high temperatures:

2Cu+O_2\rightarrow 2CuO

In such a way, for 0.032 grams of copper, the following grams of copper (II) oxide (black product) are yielded:

m_{CuO}=0.032gCu*\frac{1molCu}{63.546gCu} *\frac{2molCuO}{2molCu}*\frac{79.546gCuO}{1molCuO}  =0.078gCuO

Therefore, the percent yield is:

Y=\frac{0.038g}{0.078g}*100\%\\ \\Y=48.6\%

Best regards.

6 0
3 years ago
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