Bainliest. Mole Conversion

Rank the members of each set of compounds in order of decreasing ionic character of their bonds. Use partial charges to indicate

Firdavs [7]

Answer:

Ionic character

A. PF₃ > PBr₃ > PCl₃

B. BF₃ > CF₄ > NF₃

C. TeF₄ > BrF₃ > SeF₄

Explanation:

The most electronegative element is fluorine, followed chlorine, phosphorous nitrogen etc.

Atoms with high electronegativity tend to form negative ions.

Ionic compounds formed between elements with high electronegativity difference.

% ionic character is directly proportional to electronegativity difference.
According to Pauling Scale E.n for F(4.0), O(3.5), N(3.0), C(2.5), B(2.0), P(2.19), Se(2.55) , Te (2.1), Cl(3.16) and Br(2.96)
ΔE.N (Electronegativity difference) between( P and F = 4 - 2.19 = 1.81), (P and Br = 2.96 - 2.19 = 0.77) , (P and Cl = 3.16 - 2.96 = 0.2 )

ΔE.N (Electronegativity difference) between( N and F = 4 - 3 = 1), (B and F = 4 - 2 = 2) , (C and F = 4 - 2.5 = 1.5 )

ΔE.N (Electronegativity difference) between( Se and F = 4 - 2.55 = 1.45), (F and Te = 4 - 2.1 = 1.9) , (F and Br = 4 - 2.19 = 1.81 )

6 0

3 years ago

Boiling and melting points exploration

juin [17]

Answer:

Boiling- 212° F melting- 32°F

Explanation:

6 0

3 years ago

Hydrogen peroxide can decompose to water and oxygen by the following reaction

mestny [16]

To begin calculating, there is one thing you need to remember : $1 mole of H2O2=34.0148 g$
Then we have $5.00g of H2O2=5/34.0148=0.146995$
As you know decomposition of 2moles now has prodused 196kj
So, q is made due $0.146995 moles of H2O2$ $=(-196/2)*0.146995=-14.40551Kj$
I'm sure it will help.

6 0

2 years ago

How many grams of ag can be formed from 5.50 grams of ag2o in the equation: 2ag2o (s) → 4ag (s) o2 (g)?

oee [108]

From the given balanced equation we have find out the amount (in gm) of Ag formed from 5.50 gm of Ag₂O.

2Ag₂O(s) → 4Ag (s) + O₂ (g)

We know, molecular mass of Ag₂O= 231.7 g/mol, and atomic mass of Ag= 107.8 g/mol. Given, mass of Ag₂O=5.50 gm. Number moles of Ag₂O= $\frac{5.50}{231.7}$ = 0.0237 moles.

From the balanced chemical reaction we get 2 (two) moles of Ag₂O produces 4 (four) moles of Ag. So, 0.0237 moles of Ag₂O produces $\frac{4X0.0237}{2}$ moles=0.0474 moles of Ag= 0.0474 X 107.8 g of Ag=5.11g Ag.

Therefore, 5.50 g Ag₂O produces 5.11 g of Ag as per the given balanced chemical reaction.

4 0

3 years ago

A closed vessel system of volume 2.5 L contains a mixture of neon and fluorine. The total pressure is 3.32 atm at 0.0°C. When th

gayaneshka [121]

Answer:

moles Ne = 0.154 mol

moles F₂ = 0.217 mol

Explanation:

Step 1: Data given

Volume of the vessel system = 2.5 L

Total pressure = 3.32 atm at 0.0 °C

The mixture is heated to 15.0 °C

The entropy of the mixture increases by 0.345 J/K

The heat capacity of monoatomic gas = 3/2R and that for a diatomic gas = 5/2R

Step 2: Define the gas

Neon is a monoatomic gas, composed of Ne atoms

⇒ Cv(Ne) ≅ (3/2)R

Fluorine is a diatomic gas, composed of F₂ molecules.

⇒ Cv(F₂) ≅ (5/2)R

Step 3: Calculate moles of gas

p*V = n*R*T

⇒ with p = the total pressure = 3.32 atm

⇒ with V = the total volume = 2.5 L

⇒ with n = the number of moles of gas

⇒ with R = the gas constant = 0.08206 L*atm/K*mol

⇒ with T = the temperature = 273.15 Kelvin

n(total) = p*V/RT = (3.32 atm*2.5 L)/(0.08206 L*atm/mol•K*273.15) = 0.3703 mol

Step 4: Calculate moles of Ne and F2

For one mole heated at constant volume,

∆S = Cv*ln(288.15/273.15) = 0.05346*Cv

⇒ ∆S for 0.3703 mol,

∆S = (0.3703 mol)(0.05346)Cv = 0.345 J/K

⇒ Cv = 17.43 J/mol*K for the Ne/F₂ mixture.

For pure Ne, Cv = (3/2)R = 1.5*8.314 J/mol*K = 12.471 J/mol*K

For pure F₂, Cv = (5/2)R = 2.5 * 8.314 J/mol*K = 20.785 J/mol*K

if X is the mole fraction of Ne, we can find X by:

17.43 J/mol*K = X* 12.471 J/mol*K + (1 – X) * 20.785 J/mol*K

⇒ 20.875 – 8.314 * X = 17.43

X = 0.415 , 1 – X = 0.585

moles Ne = (0.415)(0.3703 mol) = 0.154 mol

moles F₂ = (0.585)(0.3703 mol) = 0.217 mol

4 0

3 years ago