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Pie
3 years ago
11

A neutral atom in the ground state has four electrons in its fifth and outermost principal energy level. What element is it?

Chemistry
1 answer:
Zepler [3.9K]3 years ago
7 0

Answer:

Tin

Explanation:

Tin is an element which belongs to the carbon family in the periodic table. It lies in group 14 of the periodic table i.e. IV a.

The symbol of Tin is Sn. Tin has 50 electron in total. Its atomic number is 50. The 50 electrons of tin are divided in its five energy shells. The outer most or the valence shell contains 4 electrons in the ground state.

The number of electrons present per shell in tin is : 2, 8, 18, 18, 4.

The electronic configuration of tin is : [Kr] 4d¹⁰5s²5p²

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+ H₂O
trapecia [35]

Answer:

None of these are correct, because there is no way to balance this equation, but I hope these steps help you figure out your answer.

Explanation:

Count out the single amounts of elements you have on both sides of the equation. To be balanced, you need to have the exact same for each element.

Before balanced Left side.

Cl-2

O-8

H-2

Before balanced right side.

H-1

Cl-1

O-3

That means we need to increase Hydrogen, Chlorine and Oxygen on the right for sure and see how that affects the equation. You can keep adding the Coefficients until the # of elements begin to match on each side.  

(I tried to balance this equation, it doesn't work, there is too much on the reactants side for what the product is.)

8 0
3 years ago
Be(OH)2 is solid why?​
pickupchik [31]

Answer:

no it's not solid rather it's an aqueous

Explanation:

B/c Barium hydroxide is used in analytical chemistry for the titration of weak acids, particularly organic acids. Its clear aqueous solution is guaranteed to be free of carbonate, unlike those of sodium hydroxide and potassium hydroxide, as barium carbonate is insoluble in water.

8 0
2 years ago
has a standard free‑energy change of − 3.59 kJ / mol at 25 °C. What are the concentrations of A , B , and C at equilibrium if, a
Mamont248 [21]

Answer: The concentrations of A , B , and C at equilibrium are 0.1583 M, 0.2583 M, and 0.1417 M.

Explanation:

The reaction equation is as follows.

               A + B \rightarrow C

Initial :     0.3   0.4          0

Change:  -x       -x           x

Equilbm: (0.3 - x)  (0.4 - x)  x  

We know that, relation between standard free energy and equilibrium constant is as follows.

      \Delta G = -RT ln K

Putting the given values into the above formula as follows.

      \Delta G = -RT ln K

      -3.59 kJ/mol = -8.314 \times 10^{-3} kJ/mol K ln (\frac{x}{(0.3 - x)(0.4 - x)})

                x = 0.1417

Hence, at equilibrium

  •  [A] = 0.3 - 0.1417

       = 0.1583 M

  •  [B] = 0.4 - 0.1417

       = 0.2583 M

  •  [C] = 0.1417 M
5 0
3 years ago
The picture compound has which bond?
Marina CMI [18]

Answer:

something

Explanation:

4 0
3 years ago
Read 2 more answers
An airplane travels 2100 km at 1000km/hE. It encounters a wind and slows to 800 km/h E for the next 1300 km. What is the average
Deffense [45]

Answer:

The average velocity of the airplane for this trip is 1684.21 km/h

Explanation:

Average velocity is the rate of change of displacement with time. That is,

Average velocity = \frac{Displacement }{Change in time} = Δx / Δt = \frac{x2 - x1}{t2 - t1}

Now we will calculate the time taken by the airplane for the first motion before it encounters a wind.

From,

Velocity = \frac{Distance traveled}{Time taken}

Time = \frac{Distance traveled}{Velocity}

Therefore, Time = \frac{2100km }{1000km/h}

Time = 2.1h

This is the time taken before the airplane encounters a wind.

Hence, t1 = 2.1h

Now, For the time taken by the airplane when it encounters a wind

Also from,

Velocity = \frac{Distance traveled}{Time taken}

Time = \frac{Distance traveled}{Velocity}

Therefore, Time = \frac{1300km }{800km/h}

Time = 1.625h

Hence, t2 = 1.625h

Now, to calculate the average velocity

Average velocity = \frac{x2 - x1}{t2 - t1}

x1= 2100, x2= 1300, t1= 2.1h and t2= 1.625h

Hence, Average velocity = \frac{1300 - 2100}{1.625 - 2.1}

Average velocity = 1684.21 km/h

7 0
3 years ago
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