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andrey2020 [161]
4 years ago
10

A ____________ is a large body that moves around a star?

Chemistry
2 answers:
Travka [436]4 years ago
4 0

Answer:

A Planet

Explanation:

The earth for example, is a large body that orbits the sun, our local star

Makovka662 [10]4 years ago
4 0

Answer:

Planet

Explanation:

Just like Earth orbits the sun (a star).

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Help me answer the questions in the photo i am very confused
AlladinOne [14]

Answer:

The range is also the whole of R

Explanation:

8 0
3 years ago
Which of the three has the largest ei1? which of the three has the largest ? 1s22s22p63s23p64s1 1s22s22p63s23p5 1s22s22p63s23p1?
coldgirl [10]

E_i1 will be largest for 1s^22s^22p^63s^23p^5.

Explanation: Ionization energy is the energy to knock off an electron from a gaseous atom of ion. First ionization energy or E_i1 is the energy required to remove 1 loosely held electron from 1 mole of gaseous atoms to produce 1 mole of gaseous ion carrying (+)1 charge.

M(g)\rightarrow M^+(g)+e^-

The electrons are filled according to Aufbau's rule and the orbitals which are strongly held to the nucleus follows the order s>p>d>f.

Electron is released from the outermost shell that is from the electrons which are loosely held to the nucleus, this follows the pattern s.

In configurations, 1s^22s^22p^63s^23p^64s^1,1s^22s^22p^63s^23p^5\text{ and } 1s^22s^22p^63s^23p^1

The loosely held orbital is 4s, therefore electron will be lost from that easily.

Now, in 3p orbital, one configuration has 5 electrons and one has 1 electron.

The configuration having 5 electrons will be more tightly held by the nucleus because it has more electrons that the one having only 1 electron. Hence, the electron will be lost easily from the configuration having 3p^1 as the valence shell.

Therefore, the configuration 1s^22s^22p^63s^23p^5 will the largest E_i1.


8 0
4 years ago
How many Liters of 0.968M solution can be made if 0.581 moles of solute are added? Group of answer choices 0.600 L 60 mL 0.562 L
Rzqust [24]

Answer:

0.6L

Explanation:

The formula of molarity is molSolute/litreSolution

0.968M=\frac{0.581}{LitreSolution} \\\\LitreSolution=\frac{0.581}{0.968} \\LitreSolution=0.6L.

7 0
4 years ago
All of the following explains why solar energy a better source of energy except:
san4es73 [151]

Answer:

Solar energy can be mined with less damage to ecosystems

Explanation:

solar energy isint mined, its collected and it is kinda bad for the environment when you mine the metals to get the solar panel

3 0
3 years ago
Complete and balance the following equation: S(s)+HNO3(aq)→H2SO3(aq)+N2O(g)(acidic solution)
meriva

Answer:- The balanced equation is, 2HNO_3(aq)+2S(s)+H_2O(l)\rightarrow N_2O(g)+2H_2SO_3(aq) .

Solution:- Oxidation number of S is increasing from 0 to 4 and so it is oxidation. Oxidation number of N is decreasing from 5 to 1 and so it is reduction.

We write the oxidation and reduction half equations and balance them. The given reaction is taking place in an acidic medium.

First of all we balance all the atoms other than H and O. Then oxygen is balanced by adding H_2O and hydrogen is balanced by adding H^+ . Charge is balanced by adding electrons.

To makes the electrons equal for both the half equations we multiply the equation/equations by appropriate numbers.

Oxidation half equation:

S(s)\rightarrow H_2SO_3(aq)

S is already balanced. To balance O, we need to add three water molecules to the left side:

S(s)+3H_2O(l)\rightarrow H_2SO_3(aq)

For balancing hydrogen, we need to add 4 hydrogen ions to the right side:

S(s)+3H_2O(l)\rightarrow H_2SO_3(aq)+4H^+(aq)

Now to balance the charge we need to add 4 electrons to the right side:

S(s)+3H_2O(l)\rightarrow H_2SO_3(aq)+4H^+(aq)+4e^-

Reduction half equation:

HNO_3(aq)\rightarrow N_2O(g)

To balance N, we need to multiply left side by 2:

2HNO_3(aq)\rightarrow N_2O(g)

For balancing oxygen, we need to add 5 water molecules to the right side:

2HNO_3(aq)\rightarrow N_2O(g)+5H_2O(l)

To balance hydrogen, we need to add 8 hydrogen ions to the left side:

2HNO_3(aq)+8H^+(aq)\rightarrow N_2O(g)+5H_2O(l)

Now, for charge balance, we need to add 8 electrons to the left side:

2HNO_3(aq)+8H^+(aq)+8e^-\rightarrow N_2O(g)+5H_2O(l)

First half equation has 4 electrons and second half equation has 8 electrons.

To make the electrons equal, we need to multiply oxidation half equation by 2:

2S(s)+6H_2O(l)\rightarrow 2H_2SO_3(aq)+8H^+(aq)+8e^-

Now we add both of these two half equations and cancel out common species. What we get on doing this is:

2HNO_3(aq)+2S(s)+H_2O(l)\rightarrow N_2O(g)+2H_2SO_3(aq)



6 0
3 years ago
Read 2 more answers
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