Which element has similar properties to neon (Ne)?

Nitric oxide reacts with chlorine to form nocl. the data refer to 298 k. 2no(g) + cl2(g) → 2nocl(g) substance: no(g) cl2(g) nocl

tigry1 [53]

Answer:

- 10.555 kJ/mol.

Explanation:

∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.

Where, ∆G°rxn is the standard free energy change of the reaction (J/mol).

∆H°rxn is the standard enthalpy change of the reaction (J/mol).

T is the temperature of the reaction (K).

∆S°rxn is the standard entorpy change of the reaction (J/mol.K).

Calculating ∆H°rxn:

∵ ∆H°rxn = ∑∆H°products - ∑∆H°reactants

∴ ∆H°rxn = (2 x ∆H°f NOCl) - (1 x ∆H°f Cl₂) - (2 x ∆H°f NO) = (2 x 51.71 kJ/mol) - (1 x 0) - (2 x 90.29 kJ/mol) = - 77.16 kJ/mol.

Calculating ∆S°rxn:

∵ ∆S°rxn = ∑∆S°products - ∑∆S°reactants

∴ ∆S°rxn = (2 x ∆S° NOCl) - (1 x ∆S° Cl₂) - (2 x ∆S° NO) = (2 x 261.6 J/mol.K) - (1 x 223.0 J/mol.K) - (2 x 210.65 J/mol.K) = - 121.1 J/mol.K. = - 0.1211 kJ/mol.K.

Calculating ∆G°rxn:

∵ ∆G°rxn = ∆H°rxn - T∆S°rxn.

∴ ∆G°rxn = ∆H°rxn - T∆S°rxn = (- 77.16 kJ/mol) - (550 K)(- 0.1211 kJ/mol.K) = - 10.555 kJ/mol.

4 0

3 years ago

Which property of water causes water drops to bead on a freshly waxed car?

astraxan [27]

Answer:

Cohesive forces are greater than adhesive forces

Step-by-step explanation:

The attractive forces between water molecules and the wax on a freshly-waxed car (adhesive forces) are quite weak.

However, there are strong attractive forces (cohesive forces) between water molecules.

The water molecules are only weakly attracted to the wax, so the cohesive forces pull the water molecules together to form beads .

3 0

3 years ago

For the reaction: CH3NH2(aq) + H2O(aq) ⇌ CH3NH3 +(aq) + OH- Determine the change in the pH (ΔpH) for the addition of 6.7 M CH3NH

Korolek [52]

Answer:

The change in the pH (ΔpH) is 2,17

Explanation:

The reaction:

CH₃NH₂(aq) + H₂O(aq) ⇌ CH₃NH₃⁺(aq) + OH⁻

$kb = \frac{[OH^{-}][CH_{3}NH_{3}^+]}{[CH_{3}NH_{2}]}$ (1)

In equilibrium, a solution of CH₃NH₂ 4,7M produces:

[CH₃NH₂] = 4,7 - x

[CH₃NH₃⁺] = x

[OH⁻] = x

Replacing in (1):

$4,38x10^{-4} = \frac{x^2}{4,7-x}$

x² + 4,38x10⁻⁴x - 2,0586x10⁻³ = 0

The solutions are:

x = -0,0456 No physical sense. There are not negative concentrations.

x = 0,04515 Real answer.

The concentration of [OH⁻] is 0,04515 M.

As pOH = -log [OH⁻] And pH+pOH = 14. The pH of this solution is:

pH = 12,65

The addition of 6,7M produce this changes in concentrations:

[CH₃NH₂] = 4,656 + x

[CH₃NH₃⁺] = 6,74515 - x

[OH⁻] = 0,04515 - x

Replacing in (1) you will obtain:

x² - 6,7907x + 0,3025 = 0

Solving for x:

x = 6,74586 No physical sense

x = 0,04484 Real answer.

Thus, [OH⁻] = 0,04515 - 0,044842 = 3,08x10⁻⁴M

pOH = 3,51.

pH = 10,49

Thus ΔpH is 12,65 - 10,49 = 2,16 ≈ 2,17

I hope it helps!

4 0

3 years ago

Of the choices below, which is true for the relationship shown? it is ka for the acid h3p2o72â. it is kb for the acid h3p2o72â.

Semenov [28]

A For PLATO. just did it

3 0

3 years ago

Read 2 more answers

A certain substance X has a normal freezing point of -6.4 C and a molal freezing point depression constant Kf= 3.96 degrees C.kg

Brut [27]

Answer: $1.0\times 10^2g$

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=(-6.4-(13.6))^0C=7.2^0C$ = Depression in freezing point

i= vant hoff factor = 1 (for non electrolyte like urea)

$K_f$ = freezing point constant = $3.96^0C/m$

m= molality

$\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}}\times \text{weight of solvent in kg}}$

Weight of solvent (X)= 950 g = 0.95 kg

Molar mass of non electrolyte (urea) = 60.06 g/mol

Mass of non electrolyte (urea) added = ?

$7.2=1\times 3.96\times \frac{xg}{60.06 g/mol\times 0.95kg}$

$x=1.0\times 10^2g$

Thus $1.0\times 10^2g$ urea was dissolved.

8 0

3 years ago