The density of a 50% solution of naoh is 1.525 g/ml. what volume of a solution that is 50% by weight naoh is required to make 0.

Response:

The pKa value is 13.0.

Clarification:

pKa + pKb = 14

For trimethylamine, Kb = 6.3 ×

Calculating pKb: pKb = - log (6.3 × )

= 1.0

Thus, pKa = 14 - pKb = 14 - 1.0

pKa = 13.0

Verification: The typical range for pKa in weak acids is from 2 to 13.

According to the Law, the variation in internal energy (U) of the system is equal to the total of the heat added to the system (q) plus the work performed ON the system (W)
<span>ΔU = q + W </span>

<span>In response to the first question, 0.653 kJ of heat energy is extracted from the system (balloon) while 386 J of work is applied to the balloon, leading to </span>
<span>ΔU = -653J + 386J </span>
<span>=-267J </span>
<span>Thus, the internal energy reduces by 267 J </span>

<span>For the second question, 322 J of heat is supplied to the system (gold bar) while no work is undertaken on the gold bar, marking this as an isochoric/isovolumetric process, thus </span>
<span>ΔU = 322J + 0 </span>
<span>=322J </span>
<span>Hence, internal energy rises by 322 J</span>

Answer:

Explanation:

This scenario is unrealistic since Al(OH)₃ is not soluble in water.

The question consists of two parts:

A. Stoichiometry — where we determine volumes, masses, and moles for the products

B. Calorimetry — where we assess the enthalpy of the reaction.

A. Stoichiometry

1. Determine the volume of Al(OH)₃

(a) The balanced chemical equation:

                 2Al(OH)₃ + 3H₂SO₄ ⟶ Al₂(SO₄)₃ + 6H₂O

M/V:            66.667

c/mol·L⁻¹:   4.000       3.000

(b) Moles of H₂SO₄

(c) Moles of Al(OH)₃

The molar ratio stands at 2 mmol Al(OH)₃: 3 mmol H₂SO₄

(d) Volume of Al(OH)₃

B. Calorimetry

This reaction has two energy exchanges.

q₁ = heat from the reaction

q₂ = heat used to heat the calorimeter

 q₁ + q₂ = 0

nΔH + mCΔT = 0

Data:

Moles of Al₂(SO₄)₃ = 0.066 667 mol

C = 1.10 J°C⁻¹g⁻¹

T_initial = 22.3 °C

T_final = 24.7 °C

Calculations

(a) Mass of solution

Assume solutions are as dense as water (though not realistic).

Mass of sulfuric acid solution            =   66.667 g 

Mass of aluminium hydroxide solution =  50.000    

                                             TOTAL =  116.667 g

(b) ΔT

ΔT = T_final - T_initial = 24.7 °C - 22.3 °C = 2.4°C

(c) ΔH

This result appears nonsensical, but it is derived from your given figures.

Answer:

710.33 g NO2

Explanation:

2 C8H18 + 25 O2 → 16 CO2 + 18 H2O  

(800 g octane) / (114.2293 g C8H18/mol x (25/2)) = 87.54 mol O2 utilized for combusting octane

= 15.44 mol O2 used for generating NO2

O2 + 2NO → 2NO2

(15.44 mol O2) x (2/2) x (46.0056 g NO2/mol) = 710.33 g NO2

Result: The count of bonding electrons and non-bonding electrons amounts to (4, 18).

Explanation:

The Lewis-dot structure reveals the number of bonding and non-bonding electrons in .

Lewis-dot representation: It illustrates the valence electron count for atoms in a molecule and shows how they bond, as well as any lone pairs of electrons.

In this structure, 'Xe' is the central atom while 'F' is the terminal atom.

Xenon comprises 8 valence electrons, whereas fluorine contains 7.

The total number of valence electrons in is calculated as 8 + 2(7) = 22 electrons

From the Lewis-dot structure, we can determine

The count of electrons involved in bonding = 4

The count of electrons involved in non-bonding (lone-pairs) = 22 - 4 = 18

Thus, the bonding and non-bonding electron counts are (4, 18).

Below is the Lewis-dot structure for .