What is electron configuration of sulfer?

Answers

Answer 1
Answer: Sulfur has 16 protons and 16 electrons.
The electronic configuration of Sulfur is:
2, 8, 6        =          16 

Related Questions

Match the separation techniques with the correct descriptions. 1. Heating a solid until it passes directly from the solid phase into the gaseous phase. 2. Separating a liquid from an insoluble solid sediment by carefully pouring the liquid from the solid without disturbing the solid. 3. The process of vapor returning to the solid phase without a liquid phase in between. 4. Heating a mixture to vaporize a volatile liquid component to make the remaining component dry. 5. Separating a solid from a liquid using a porous material, such as paper, charcoal, or sand, as a filter. 6. Using a solvent to selectively dissolve one or more components from a solid mixture.A. SublimationB. Solids that SublimeC. ExtractionD. DecantationE. EvaporationF. Filtration
The breaking and making of chemical bonds can explain chemical reactions and energy flow. a. True b. False
What is the main factor that determines if a star will become a black hole
Which element is classified as an alkali metal? Rb Sr Zr Cl
The pH of a .1 M solution is 11. What is the concentration of H3O ions, in miles per liter

The compound XCl is classified as ionic if X represents the element:

Answers

If the element is a metal with one electron in its outermost energy level.

If you want to make 0.5 L of a 0.01 M solution of bromine (Br2) in water, how much bromine would you need? A. 0.8 grams. . B. 0.8 moles. . C. 1.6 grams. . D. 1.6 moles.

Answers

The answer is A. 0.8 grams.

The atomic mass of bromine is: A(Br) = 80
The molecule Br₂ consists of 2 Br atoms. Thus, the molecular mass of bromine is: Mr(Br₂) = 2*A(Br) = 2*80 = 160
This means there are 160 grams/l of Br₂ in 1 M.

Let's write a proportion. If 160 grams/l of Br₂ are present in 1 M, how much of Br₂ will be in 0.01 M:
160 g/l : 1 M = x : 0.01 M
After crossing the products:
x = 80 * 0.01 = 1.6 g/l

Let's write another proportion. If there are 1.6 g of Br₂ in 1 liter, how many of Br₂ will be present in 0.5 liter:
1.6 g : 1 l = x : 0.5
After crossing the products:
x = 1.6 * 0.5 = 0.8 g

Why does an egg sink in fresh water but float in salt water??

Answers

If an object is less dense than the water around it, it will float. Because salt water is denser than freshwater, some things float more easily in the ocean—or extremely salty bodies of the water, such as the Dead Sea. You can make your own dense water by adding salt to tap water.

Gold and helium are examples of a(n)

Answers

Gold and helium are the example of element.

What is element?

A material is considered an element if it could not be divided into smaller parts by whatever non-nuclear chemical reaction.

What is gold?

Gold is a kind of element which has atomic number 79 and can be represented by Au.

A soft, golden metal, gold is. Gold was incredibly malleable as well as ductile, along with all other metals. Gold also has a good ability to conduct heat as well as electricity.

To know more about element and gold.

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gold and helium are examples of elements

How many grams of iron metal do you expect to be produced when 265 grams of an 84.5 percent by mass iron (II) nitrate solution react with excess aluminum metal? Show all of the work needed to solve this problem. 2Al (s) + 3Fe(NO3)2 (aq) yields 3Fe (s) + 2Al(NO3)3 (aq)

Answers

The balanced reaction is:

2Al (s) + 3Fe(NO3)2 (aq) = 3Fe (s) + 2Al(NO3)3 (aq)

We are given the amount of iron (II) nitrate solution with its purity. This will be the starting point of our calculation.

265 (.845) gram 
iron (II) nitrate (1 mol Fe(NO3)2 / 179.85 g Fe(NO3)2) ( 3 mol Fe / 3 mol Fe(NO3)2) (55.85 g Fe / 1 mol Fe) = 69.54 g Fe

The standard cell potential for the aqueous reduction of Pb4+ to Pb2 + ions by the corresponding oxidation of Ce3+ to Ce4+ ions, Pb4 •(aq) + 2 Ce' •(aq) --'> Pb'•(aq) + 2 Ce4+(aq) is 0.06 V a

Answers

Answer:the standard reduction potential (E°red) for the reduction half-reaction of Pb^4+(aq) to Pb^2+(aq) is 1.50 V.

Explanation:The given chemical reaction is:

Pb^4+(aq) + 2 Ce^3+(aq) -> Pb^2+(aq) + 2 Ce^4+(aq)

The standard cell potential (E°cell) for this electrochemical reaction is 0.06 V.

The standard cell potential for a galvanic cell can be calculated using the Nernst equation:

E°cell = E°cathode - E°anode

In this reaction, Pb^4+(aq) is being reduced to Pb^2+(aq), so it is the reduction half-reaction, and Ce^3+(aq) is being oxidized to Ce^4+(aq), so it is the oxidation half-reaction.

The standard reduction potentials (E°) for the half-reactions are as follows:

For the reduction half-reaction:

Pb^4+(aq) + 2 e^- -> Pb^2+(aq) E°red = x (we'll solve for x)

For the oxidation half-reaction:

2 Ce^3+(aq) -> 2 Ce^4+(aq) + 2 e^- E°red = 1.44 V (This value is usually given)

Now, plug these values into the Nernst equation:

E°cell = E°cathode - E°anode

0.06 V = x - 1.44 V

Now, solve for x:

x = 0.06 V + 1.44 V

x = 1.50 V

So, the standard reduction potential (E°red) for the reduction half-reaction of Pb^4+(aq) to Pb^2+(aq) is 1.50 V.