Explain why the actual electron configurations of 2-7-1-1 represents a sodium in an excited state

What is the pH of a solution made by mixing 15.00 mL of 0.100 M HCl with 50.00 mL of 0.100 M KOH? Assume that the volumes of the

denis23 [38]

Answer:

The correct answer is: pH = 12.73

Explanation:

The <em>neutralization reaction</em> between HCl and KOH is given by the following chemical equation:

HCl + KOH ⇒ KCl + H₂O

Since HCl is a strong acid and KOH is a strong base, HCl is completely dissociated into H⁺ and Cl⁻ ions, whereas KOH is dissociated completely into K⁺ and OH⁻ ions.

For acids, the number of equivalents is given by the moles of H⁺ ions (in this case: 1 equivalent per mol of HCl). For bases, the number of equivalents is given by the moles of OH⁻ ions (in this case: 1 equivalent per mol of KOH).

The H⁺ ions from HCl will react with OH⁻ ions of KOH to give H₂O. The pH is calculated from the difference between the equivalents of H⁺ and OH⁻:

equivalents of H⁺= volume HCl x Molarity HCl

= (15.0 mL x 1 L/1000 mL) x 0.100 mol/L

= 1.5 x 10⁻³ eq H⁺

equivalents of OH⁻= volume KOH x Molarity KOH

= (50.0 mL x 1 L/1000 mL) X 0.100 mol/L

= 5 x 10⁻³ eq OH⁻

There are more OH⁻ ions than H⁺ ions. The excess of OH⁻ (that did not react with H⁺ ions) is calculated as follows:

OH⁻ ions= (5 x 10⁻³ eq OH⁻) - (1.5 x 10⁻³ eq H⁺) = 3.5 x 10⁻³ eq OH⁻= 3.5 x 10⁻³ moles OH⁻

As the volumes of the solutions are additive, the total volume of the solution is:

V= 15.0 mL + 50.0 mL = 65.0 mL= 0.065 L

So, the concentration of OH⁻ ions in the solution is given by:

[OH⁻] = moles OH⁻/V= (3.5 x 10⁻³ moles OH⁻)/0.065 L = 0.054 mol/L = 0.054 M

From [OH⁻], we can calculate pOH:

pOH = -log [OH⁻] = -log (0.054) = 1.27

Finally, we know that pH + pOH= 14; so we calculate pH:

pH= 14 - pOH = 14 - 1,27 = 12.73

8 0

2 years ago

Two nonpolar organic liquids, hexane (C6H14) and heptane (C7H16), are mixed. Do you expect ΔHsoln to be a large positive number,

Aliun [14]

Answer:

ΔH of solution is expected to be close to zero.

Explanation:

When we mix two non polar organic liquids like hexane and heptane,the resulting mixture formed is an ideal solution.An ideal solution is formed when the force of attraction between the molecules of the two liquids is equal to the force of attraction between the molecules of the same type.

For instance if liquids A and B are mixed,

$F_{A-A}$ = $F_{B-B}$ = $F_{A-B}$

Hence the condition before and after mixing remains unchanged.

Since enthalpy change is associated with inter molecular force of attraction the enthalpy change for ideal solution is zero.

More examples of ideal solutions are:

1. Ethanol and Methanol

2. Benzene and Toluene

3. Ethyl bromide and Ethyl iodide

5 0

1 year ago

If a large marshmallow has a volume of 2.50 i n 3 and density of 0.242 g/c m 3 , how much would it weigh in grams? 1 i n 3 =16.3

Illusion [34]

Density of a substance is defined as the mass of the substance divided by the volume.

Density of the substance= 0.242 g cm⁻³

volume of the substance= 2.50 in³

As, 1 in³= 16.39 cm³

So, 2.50 in³= 16.39× 2.50 cm³=40.97 cm³

As ,

$Density=\frac{mass}{volume}$

Mass=volume ×Density

Mass=40.97 × 0.242

Mass=9.916 g.

4 0

2 years ago

What is the maximum volume of a 0.788 M CaCl2 solution that can be prepared using 85.3 g CaCl2?

Anna11 [10]

Molar mass CaCl₂ = 110.98 g/mol

Number of moles:

1 mole CaCl₂ ---------> 110.98 g
n mole CaCl2 ---------> 85.3 g

n = 85.3 / 110.98

n = 0.7686 moles of CaCl₂

Volume = ?

M = n / V

0.788 = 0.7686 / V

V = 0.7686 / 0.788

V = 0.975 L

hope this helps!

5 0

2 years ago

Three of the reactions that occur when the paraffin of a candle (typical formula C21H44) burns are as follows:

Katarina [22]

Answer and Explanation:

The 3 reactions represented are

C₂₁H₄₄ + 32O₂ -----> 21CO₂ + 22H₂O

C₂₁H₄₄ + (43/2)O₂ -----> 21CO + 22H₂O

C₂₁H₄₄ + 11O₂ -----> 21C + 22H₂O

ΔH°(C₂₁H₄₄) = -476 KJ/mol, ΔH°(O₂) = 0KJ/mol, ΔH°(CO₂) = -393.5 KJ/mol, ΔH°(CO) = -99 KJ/mol, ΔH°(H₂O) = -292.74 KJ/mol, ΔH°(C) = 0KJ/mol

ΔH°f = ΔH°(products) - ΔH°(reactants)

For reaction 1,

ΔH°(products) = 21(ΔH°(CO₂)) + 22(ΔH°(H₂O)) = 21(-393.5) + 22(-292.74) = -14703.78 KJ/mol

ΔH°(reactants) = ΔH°(C₂₁H₄₄) = -476 KJ/mol

ΔH°f = -14703.78 - (-476) = - 14227.78 KJ/mol

For reaction 2,

ΔH°(products) = 21(ΔH°(CO)) + 22(ΔH°(H₂O)) = 21(-99) + 22(-292.74) = -8519.28 KJ/mol

ΔH°(reactants) = ΔH°(C₂₁H₄₄) = -476 KJ/mol

ΔH°f = -8519.28 - (-476) = - 8043.28 KJ/mol

For reaction 3,

ΔH°(products) = 21(ΔH°(C)) - 22(ΔH°(H₂O)) = 21(0) + 22(-292.74) = -6440.28 KJ/mol

ΔH°(reactants) = ΔH°(C₂₁H₄₄) = -476 KJ/mol

ΔH°f = -6440.28 - (-476) = - 5968.28 KJ/mol

b) To obtain the q for the combustion of 254g of paraffin, we convert the mass to moles.

Number of moles = mass/molar mass; molar mass of C₂₁H₄₄ = 296 g/mol

Number of moles = 254/296 = 0.858 mole

heat of reaction for the combustion of C₂₁H₄₄ when it is complete combustion, q = ΔH°(complete combustion, i.e. reaction 1) × number of moles = -14227.78 × 0.858 = -12207.435 KJ/mol

c) 8% of the mass of C₂₁H₄₄ undergoes incomplete combustion = 8% × 254 = 20.32g, in number of moles = 20.32/296 = 0.0686 mole

5% of the mass of C₂₁H₄₄ becomes soot = 5% × 254 = 12.7g, in number of moles = 12.7/296 = 0.0429 mole

The remaining paraffin undergoes complete combustion = 87% of 254 = 220.98g, in number of moles = 220.98 = 0.747 mole

q = sum of all the heat of reactions = (0.747 × -14227.78) + (0.0686 × -8043.28) + ( 0.0429 × -5968.28) = -11435.377 KJ

QED!!!

8 0

1 year ago