All categories

2 years ago

Which species does not have a noble gas electron

Chemistry

2 answers:

Inessa [10]2 years ago

8 0

Noble gases are elements with complete octet configuration

this means their outershells are complete with 8 electrons

other elements have incompletely filled outershells therefore they either lose or gain electrons to gain a complete octet in the outershell

1. Na⁺ - Na electron configuration is 2,8,1

positive charge means it loses its outermost shell electron and becomes 2,8 with 8 electrons in the outermost shell

therefore has a noble gas configuration

2.Mg²⁺ - Mg electron configuration is 2,8,2

2 plus means Mg loses 2 electrons in the outermost shell and becomes 2,8 and has 8 electrons in the outer energy shell

therefore has a noble gas configuration

3.Ar - Ar electron configuration is 2,8,8

Ar has 8 electrons in the outermost energy shell therefore its complete

Ar element is a noble gas

4.S - S electron configuration is 2,8,6

S has 6 electrons in the outermost energy

therefore it does not have a noble gas configuration

correct answer is 4) S

lara31 [8.8K]2 years ago

6 0

S, sulfur does not have a noble gas electron.

You might be interested in

Determine the compound type for the following formulas: C12H22011 Mg(OH)2 H20 Cu3Zn2 Au <br>

Scorpion4ik [409]

Answer:

C12H22O11

✔ covalent

Mg(OH)2

✔ Ionic

H2O

✔ covalent

Cu3Zn2

✔ metallic

Explanation:

7 0

2 years ago

Read 2 more answers

A vessel of capacity 400 cc filled with chlorine under 80 cm is connected by a narrow tube and stopcock with another vessel of c

guapka [62]

Answer:

The pressure when the stopcock is opened is opened is 87.783 cm

Explanation:

The given parameters of the question are;

The volume of the vessel of chlorine = 400 cc

The pressure of the vessel of chlorine = 80 cm

The volume of the vessel of chlorine = 250 cc

The pressure of the vessel of chlorine = 100 cm

Daltons law of Partial Pressures states that the total pressure exerted by a volume of a mixture of gases is equal to the partial pressures exerted by the individual gases in the mixture with respect to the given volume

Therefore;

The total volume of the mixture = 400 cc + 250 cc = 650 cc

The partial pressure exerted by the chlorine gas in the total volume is given by Boyles law as follows;

P₁·V₁ = P₂·V₂

P₂ = P₁·V₁/V₂

Where;

P₁ = 80 cm = The pressure in volume V₁ = 400 cc

P₂ₓ = The partial pressure of chlorine in volume V₂ = 650 cc

Substituting, we have;

P₂ₓ = 80 × 400/650 ≈ 49.321 cm

Similarly, the partial pressure exerted by the nitrogen gas in the total volume is given by Boyles Law as follows;

P₂ₐ = P₁·V₁/V₂

Where;

P₁ = 100 cm = The pressure in volume V₁ = 250 cc

P₂ₐ = The partial pressure of nitrogen in volume V₂ = 650 cc

Substituting, we have;

P₂ₐ = 100× 250/650 ≈ 38.462 cm

The pressure of the combined gas, P, when the stopcock is opened is opened is given by Dalstons Law of partial pressure as P = P₂ₐ + P₂ₓ

Therefore, the pressure, P when the stopcock is opened is opened = 49.321 cm + 38.462 cm = 87.783 cm

3 0

1 year ago

A student carries out the precipitation reaction shown below, starting with 0.030 moles of calcium nitrate. The final mass of th

valkas [14]

Answer:

a. Ca₃(PO₄)₂.

b. 0.010 moles of Ca₃(PO₄)₂ can we expect to be produced

c. 3.1g of Ca₃(PO₄)₂

d. Percent yield = 93.5%

Explanation:

a. Based on the reaction:

3Ca(NO₃)₂(aq) + 2Na₃PO₄(aq) → Ca₃(PO₄)₂(s) + 6NaNO₃(aq)

<em>3 moles of calcium nitrate reacts with 2 moles of sodium phosphate producieng 1 mole of calcium phosphate.</em>

<em />

As you can see, Ca₃(PO₄)₂ is a solid product -(s)-, that means when the reaction occurs the precipitate produced is the solid,

b. As 3 moles of calcium nitrate produce 1 mole of calcium phosphate and there are 0.030 moles of calcium nitrate

0.030 moles Ca(NO₃)₂ × (1 mol Ca₃(PO₄)₂ / 3 moles Ca(NO₃)₂) =

<h3>0.010 moles of Ca₃(PO₄)₂ can we expect to be produced</h3><h3 />

c. As molar mass of Ca₃(PO₄)₂ is 310.18g/mol, the mass of 0.010 moles (The expected mass) is;

0.010 moles Ca₃(PO₄)₂ × (310.18g / mol) =

d. The percent yield is defined as 100 times the ratio between the obtained yield (That is 2.9g of precipitate, Ca₃(PO₄)₂) and the expected yield, 3.1g of Ca₃(PO₄)₂:

$\frac{2.9g}{3.1g} *100$

<h3>Percent yield = 93.5%</h3>

7 0

2 years ago

A 12.0 g sample of a metal is heated to 90.0 ◦C. It is then dropped into 25.0 g of water. The temperature of the water rises fro

Liula [17]

Answer:

The specific heat of the metal is 0.335 J/g°C

Explanation:

<u>Step 1:</u> Data given

Mass of the metal = 12.0 grams

Initial temperature of the metal = 90.0 °C

Mass of the water = 25.0 grams

Initial temperature of water = 22.5 °C

Final temperature of water (and metal) = 25.0 °C

The specific heat of water = 4.18 J/g°C

<u>Step 2:</u> Calculate the specific heat of the metal

Qgained = -Qlost

Qwater = -Qmetal

Q= m*c*ΔT

m(metal) *c(metal)*ΔT(metal) = -m(water)*c(water)*ΔT(water)

⇒ mass of the metal = 12.0 grams

⇒ c(metal) = TO BE DETERMINED

⇒ ΔT(metal) = T2 - T1 = 25.0 - 90.0 °C = -65.0

⇒ mass of the water = 25.0 grams

⇒ c(water) = the specific heat of water = 4.18 J/g°C

⇒ ΔT(water) = T2 - T1 = 25.0 - 22.5 = 2.5°C

12.0 * c(metal) * -65.0 = -25.0 * 4.18 * 2.5

c(metal) = 0.335 J/g°C

The specific heat of the metal is 0.335 J/g°C

6 0

1 year ago

How many grams of the titrant used in the question 3 (sodium hydroxide -- 0.1001 mole/kg of solution) are required to titrate 10

katen-ka-za [31]

846.6 g of sodium hydroxide solution

Explanation:

The chemical reaction of butanedioic acid with sodium hydroxide:

HOOC-CH₂-CH₂-COOH + 2 NaOH → NaOOC-CH₂-CH₂-COONa + 2 H₂O

To find the mass of the solute (butanedioic acid) from a solution about which we know that the weight/weight percent concentration, we use the following formula:

concentration / 100 = solute mass / solution mass

solute mass = (concentration × solution mass) / 100

solute mass (butanedioic acid mass) = (5 × 10) / 100 = 0.5 g

number of moles = mass / molar weight

number of moles of butanedioic acid = 0.5 / 118 = 0.04237 moles

Knowing the chemical reaction we devise the following reasoning:

if 1 mole of butanoic acid reacts with 2 moles of sodium hydroxide

then 0.04237 moles of butanoic acid reacts with X moles of sodium hydroxide

X = (0.04237 × 2) / 1 = 0.08474 moles of sodium hydroxide

If the concentration of the titrant, sodium hydroxide solution, is 0.1001 moles / 1000 g of solution the we devise the following reasoning:

if there are 0.1001 moles of sodium hydroxide in 1000 g of solution

then there are 0.08474 moles of sodium hydroxide in Y g of solution

Y = (0.08474 × 1000) / 0.1001 = 846.6 g of solution

Learn more about:

weight/weight percent concentration

brainly.com/question/3830901

#learnwithBrainly

5 0

1 year ago