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frutty [35]
4 years ago
10

How many orbitals are there with a principal quantum number of 1?

Chemistry
1 answer:
Tema [17]4 years ago
8 0

Answer:

1 orbal.

Explanation:

There is only one orbital in the n = 1 shell because there is only one way in which a sphere can be oriented in space. The only allowed combination of quantum numbers for which n = 1 is the following. There are four orbitals in the n = 2 shell.

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dexar [7]

Energy is stored in chemical bonds during photosynthesis.

During photosynthesis, the radiant energy from the sun is converted to chemical energy in carbohydrates.

Inorganic materials in the form of carbon dioxide and oxygen combine to form carbohydrates in the presence of radiant energy according to the equation below:

6CO_2 + 6H_2O ---> C_6H_1_2O_6 + 6O_2

The energy is thus, stored in chemical bonds in the carbohydrate and this is what is oxidized during respiration to release the locked energy.

More on photosynthesis can be found here: brainly.com/question/1388366

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2 years ago
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If half of the moon is always illuminated why does its appearance from earth change?
irina [24]

Answer:

The different angles the moon has from sunlight to the earth

Explanation:

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3 years ago
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Solid silicon and solid magnesium chloride form when silicon tetrachloride gas reacts with magnesium metal. Write a word equa- t
antiseptic1488 [7]
SiCl4 + 2Mg ---> 2Si + 2MgCl2
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3 years ago
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Yuri [45]

Answer:

Qura'an is what makes my day, I read it every morning and it makes me relaxed. I love reading the Qura'an. Do dua for me because I hope I am gonna memorize it after I finish reading it with tajweed.

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How much energy must be removed from a 94.4 g sample of benzene (molar mass= 78.11 g/mol) at 322.0 K to solidify the sample and
Kay [80]

Answer : The energy removed must be, 29.4 kJ

Explanation :

The process involved in this problem are :

(1):C_6H_6(l)(322K)\rightarrow C_6H_6(l)(279K)\\\\(2):C_6H_6(l)(279K)\rightarrow C_6H_6(s)(279K)\\\\(3):C_6H_6(s)(279K)\rightarrow C_6H_6(s)(205K)

The expression used will be:  

Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+[m\times \Delta H_{fusion}]+[m\times c_{p,s}\times (T_{final}-T_{initial})]

where,

Q = heat released for the reaction = ?

m = mass of benzene = 94.4 g

c_{p,s} = specific heat of solid benzene = 1.51J/g^oC=1.51J/g.K

c_{p,l} = specific heat of liquid benzene = 1.73J/g^oC=1.73J/g.K

\Delta H_{fusion} = enthalpy change for fusion = -9.8kJ/mol=-\frac{9.8\times 1000J/mol}{78g/mol}=-125.6J/g

Now put all the given values in the above expression, we get:

Q=[94.4g\times 1.73J/g.K\times (279-322)K]+[94.4g\times -125.6J/g]+[94.4g\times 1.51J/g.K\times (205-279)K]

Q=-29427.312J=-29.4kJ

Negative sign indicates that the heat is removed from the system.

Therefore, the energy removed must be, 29.4 kJ

3 0
3 years ago
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