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3 years ago

How many electron shells/orbitals does the element Krypton (Kr) have?

Chemistry

1 answer:

dlinn [17]3 years ago

4 0

Answer:

Krypton has 4 orbital shells

Extra Information:

Krypton has 36 electrons which are arranged in these 4 orbit shells,

The order in which electrons are arranged in these shells is 2,8,18,8

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Changes in temperature, pressure, and volume affect the behavior of gases more than they affect solids and liquids. ____________

Marina CMI [18]

True. All of those things affect gases more.

4 0

2 years ago

The arsenic in a 1.223 g sample of a pesticide was converted to H3AsO4 by suitable treatment. The acid was then neutralized, and

Vladimir79 [104]

Answer:

5.471% As₂O₃ in the sample.

Explanation:

...the reaction is: Ag+ + SCN- => AgSCN(s) Calculate the percent As2O3 in the sample. (F.W. As2O3 = 197.84 g/mol).

First, with the amount of KSCN we can find the moles of Ag in the filtrates. As we know the amount of Ag added we can know the precipitate of Ag and the moles of AsO₄ = 1/2 moles of As₂O₃ in the sample:

Moles KSCN = Moles Ag⁺ in the filtrate:

0.01127L * (0.100mol / L)= 0.001127moles Ag⁺

Total moles Ag⁺:

0.0400L * (0.0781mol/L) = 0.0031564 moles Ag⁺

Moles of Ag⁺ in the precipitate:

0.0031564 - 0.001127 = 0.0020294 moles Ag⁺

Moles AsO₄ = Moles As:

0.0020294 moles Ag⁺ * (1mol As / 3 moles Ag⁺) = 6.765x10⁻⁴ moles AsO₄

Moles As₂O₃:

6.765x10⁻⁴ moles AsO₄ * (1 mol As₂O₃ / 2 mol AsO₄) =

3.382x10⁻⁴ moles As₂O₃

Mass As₂O₃:

3.382x10⁻⁴ moles As₂O₃ * (197.84g/mol) = 0.0669g As₂O₃

Percent is:

0.0669g As₂O₃ / 1.223g sample * 100 =

<h3>5.471% As₂O₃ in the sample</h3>

7 0

3 years ago

A sample of 23.2 g of nitrogen gas is reacted with

slavikrds [6]

Answer:

1.66 moles.

Explanation:

We'll begin by calculating the number of mole in 23.2 g of nitrogen gas, N2.

This is illustrated below:

Molar mass of N2 = 2x14 = 28 g/mol

Mass of N2 = 23.2 g

Mole of N2 =.?

Mole = mass /Molar mass

Mole of N2 = 23.2/28

Mole of N2 = 0.83 mole

Next, we shall determine the number of mole in 23.2 g of Hydrogen gas, H2.

This is illustrated below:

Molar mass of H2 = 2x1 = 2 g/mol

Mass of H2 = 23.2 g

Mole of H2 =?

Mole = mass /Molar mass

Mole of H2 = 23.2/2

Mole of H2 = 11.6 moles

Next, the balanced equation for the reaction. This is given below:

N2 + 3H2 —> 2NH3

From the balanced equation above,

1 mole of N2 reacted with 3 moles of H2 to produce 2 moles of NH3.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

1 mole of N2 reacted with 3 moles of H2.

Therefore, 0.83 moles will react with = (0.83 x 3) = 2.49 moles of H2.

From the calculations made above, we can see that only 2.49 moles out of 11.6 moles of H2 is required to react completely with 0.83 mole of N2.

Therefore, N2 is the limiting reactant.

Finally, we shall determine the maximum amount of NH3 produced from the reaction.

In this case, we shall use the limiting reactant because it will give the maximum yield of NH3 since all of it is consumed in the reaction.

The limiting reactant is N2 and the maximum amount of NH3 produced can be obtained as follow:

From the balanced equation above,

1 mole of N2 reacted to produce 2 moles of NH3.

Therefore, 0.83 mole of N2 will react to produce = (0.83 x 2) = 1.66 moles of NH3.

Therefore, the maximum amount of NH3 produced from the reaction is 1.66 moles.

5 0

3 years ago

consider the following equilibrium:h2co3(aq) h2o(l) h3o (aq) hco3-1(aq).what is the correct equilibrium expression?

MAVERICK [17]

The equilibrium expression shows the ratio between products and reactants. This expression is equal to the concentration of the products raised to its coefficient divided by the concentration of the reactants raised to its coefficient. The correct equilibrium expression for the given reaction is:

H2CO3(aq) + H2O(l) = H3O+(aq) + HCO3-1(aq)

Kc = [HCO3-1] [H3O+] / [H2O] [H2CO3]

4 0

3 years ago

Swim ability is it materials ability to burn in the presence of

eduard

Answer: Flammability is a material's ability to burn in the presence of oxygen.

Explanation: Chemical properties can be observed only when the substance changes into one or more different substances through chemical reactions or transformations. One of the chemical properties is flammability.

Flammability is a material's ability to burn in the presence of oxygen.

Remember, oxygen doesn't burn. Precisely flammable substances obtain substances that burn. Oxygen remains an oxidizing agent, which means it supports the combustion process. Oxygen causes other objects to catch fire at low temperatures and burns hotter and faster. But oxygen itself does not burn. Consequently, if you at present deliver fuel and fire, adding oxygen will provide the fire.

Carbon dioxide is the result of combustion. An example can be seen in firewood in a fireplace. One of the chemical properties of carbon-based wood is having the ability to burn. Chemically the wood turns into carbon dioxide when it burns and leaves a residue of ash. Furthermore, this ash residue cannot be turned back into the wood. Chemical changes result in new substances.

Consider an example of a combustion reaction to methane gas:

Our balanced equation for methane combustion implies that every one CH₄ molecule reacts with two O₂ molecules. The product of combustion is one carbon dioxide molecule and two steam or water vapor molecules.

6 0

3 years ago