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zlopas [31]
3 years ago
12

Ordinary hydrogen is sometimes called a perfect fuel, because of its almost unlimited supply on Earth, and when it burns, harmle

ss water is the product of the combustion. So why don't we abandon fission energy and fusion energy, not to mention fossil-fuel energy, and just use hydrogen?
Chemistry
1 answer:
Mrac [35]3 years ago
8 0

Answer:

This is because Hydrogen represents a viable solution for stored energy but not as a general energy source since obtaining the hydrogen from water

Explanation:

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Draw a picture of the moon and sun illustrating the full moon phrase
jek_recluse [69]

Hi Sydney!

I can't draw in this question, but there is a picture showing this phase for you to follow when you draw it.

Hope This Helps :)

6 0
4 years ago
Read 2 more answers
Describe the properties of alkali metals. Based on their electronic arrangement, explain whether they exist alone in nature.
bearhunter [10]

Answer:

- They are highly reactive metals

- They have low electro negativity

- They have low ionization energy

- They don't exist alone in nature

- They have low densities

Explanation:

Alkali metals are the elements in group 1 of the periodic table. They include Sodium, Lithium, Potassium e.t.c.

Due to the fact they have one atom in their outermost shell, they are very unstable because they easily react with other elements and are therefore don't exist alone in nature but combined with other elements for this same reason.

Since alkali metals don't easily attract other elements due to it's lone pair in the outer most shell, it can be said to have low electro negativity.

Also, they don't need energy to discharge their electrons since they are highly reactive due to their lone pair in the outermost shell and so we say they have low ionization energy.

Due to this reason, they also have low densities.

7 0
3 years ago
Which element would you expect to be more reactive: phosphorus (P) or fluorine (F)
Studentka2010 [4]

Answer:

Fluorine

Explanation:

Fluorine is the most electronegative element because it has 5 electrons in it's 2P shell.

4 0
3 years ago
Read 2 more answers
For the cell constructed from the hydrogen electrode and metal-insoluble salt electrode, B) calculate the mean activity coeffici
Brut [27]

<u>Question:</u>

For the cell constructed from the hydrogen electrode and metal-insoluble salt electrode, B) calculate the mean activity coefficient for 0.124 b HCl solution if E=0.342 V at 298 K

<u>Answer:</u>

The mean activity coefficient for HCl solution is 0.78.

<u>Explanation:</u>

Activity coefficient is defined as the ratio of any chemical activity of any substance with its molar concentration. So in an electrochemical cell, we can write activity coefficient as γ. The equation for determining the mean activity coefficient is

              E=E_{0}-0.0514 \mathrm{V} \ln \gamma

As we know that E_{0} = 0.22 V and E = 0.342 V, the equation will become

             0.342 V+0.0514 V \ln (0.124)=0.22 V-0.0514 V \ln \gamma

             0.342 V-0.222 V=-0.0514 V(\ln \gamma+\ln 0.124)

             0.12 \mathrm{V}=-0.0514 \mathrm{V}(\ln \gamma+\ln 0.124)

             \frac{0.12}{0.0514}=-\ln (0.124 \gamma)

             -2.3346=\ln (0.124 \gamma)

             e^{-2.3346}=0.124 \gamma

             0.0968=0.124 \gamma

             \gamma=\frac{0.0968}{0.124}=0.78

So, the mean activity coefficient is 0.78.

6 0
3 years ago
Calcium oxide or quicklime (CaO) is used in steelmaking, cement manufacture, and pollution control. It is prepared by the therma
Elena-2011 [213]

Answer:

The yearly release of CO_2 into the atmosphere is 6.73\times 10^{10} kg.

Explanation:

CaCO_3(s)\rightarrow CaO(s) + CO_2(g)

Annual production of CaO = 8.6\times 10^{10} kg=8.6\times 10^{13} g

Moles of CaO :

\frac{8.6\times 10^{13} g}{56 g/mol}=1.53\times 10^{12} moles

According to reaction, 1 mole of CaO is produced along with 1 mole of carbon-dioxide.

Then along with  1.53\times 10^{12} moles of CaO moles of carbon-dioxide moles produced will be:

\frac{1}{1}\times 1.53\times 10^{12} moles=1.53\times 10^{12} moles of carbon-dioxide

Mass of 1.53\times 10^{12} moles of carbon-dioxide:

1.53\times 10^{12}mol\times 44 g/mol=6.73\times 10^{13} g =6.73\times 10^{10} kg

The yearly release of CO_2 into the atmosphere is 6.73\times 10^{10} kg.

6 0
3 years ago
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