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tester [92]
2 years ago
7

Ozone, O3(g), is a form of elemental oxygen produced during electrical discharge. Is ΔH∘f for O3(g) necessarily zero? Yes or no

question?
Chemistry
2 answers:
Ivan2 years ago
6 0
The answer for your question is <span>No. This is because in given conditions, it is not the most stable form of oxygen's element. It will not equate into zero because there will be charge remained after balancing the equation. 
</span>
stiv31 [10]2 years ago
6 0

Answer: ΔH°f is not zero

Explanation:

the change in enthalpy formation is zero when elements are pure.when they are zero it is mainly  because they are most basic,where you cannot form an element with an element.

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The human body can get energy by metabolizing proteins, carbohydrates or fatty acids, depending on the circumstances. Roughly sp
Ede4ka [16]

Answer:

3:1

Explanation:

Stearic acid - C18H36O2

Fructose- C6H12O6

Since the energy content of food is roughly proportional to the carbon content. It means:

C18/C6 = 3 or 3:1

This explains why fats and oil are the major source of energy in the body and not carbohydrates. This is due to Fats and oil having larger amounts of Carbon compared to Carbohydrates.

6 0
2 years ago
A 60.0 mL solution of 0.112 M sulfurous acid (H2SO3) is titrated with 0.112 M NaOH. The pKa values of sulfurous acid are 1.857 (
djverab [1.8K]

Answer:

a)4.51

b) 9.96

Explanation:

Given:

NaOH = 0.112M

H2S03 = 0.112 M

V = 60 ml

H2S03 pKa1= 1.857

pKa2 = 7.172

a) to calculate pH at first equivalence point, we calculate the pH between pKa1 and pKa2 as it is in between.

Therefore, the half points will also be the middle point.

Solving, we have:

pH = (½)* pKa1 + pKa2

pH = (½) * (1.857 + 7.172)

= 4.51

Thus, pH at first equivalence point is 4.51

b) pH at second equivalence point:

We already know there is a presence of SO3-2, and it ionizes to form

SO3-2 + H2O <>HSO3- + OH-

Kb = \frac{[ HSO3-][0H-]}{SO3-2}

Kb = \frac{10^-^1^4}{10^-^7^.^1^7^2} = 1.49*10^-^7

[HSO3-] = x = [OH-]

mmol of SO3-2 = MV

= 0.112 * 60 = 6.72

We need to find the V of NaOh,

V of NaOh = (2 * mmol)/M

= (2 * 6.72)/0.122

= 120ml

For total V in equivalence point, we have:

60ml + 120ml = 180ml

[S03-2] = 6.72/120

= 0.056 M

Substituting for values gotten in the equation Kb=\frac{[HSO3-][OH-]}{[SO3-2]}

We noe have:

1.485*10^-^7=\frac{x*x}{(0.056-x)}

x = [OH-] = 9.11*10^-^5

pOH = -log(OH) = -log(9.11*10^-^5)

=4.04

pH = 14- pOH

= 14 - 4.04

= 9.96

The pH at second equivalence point is 9.96

4 0
2 years ago
What mass of calcium carbonate is produced when 250 mL of 6.0 M sodium carbonate is added to 750 mL of 1.0 M calcium fluoride
Savatey [412]

<u>Given:</u>

Volume of Na2CO3 = 250 ml = 0.250 L

Molarity of Na2CO3 = 6.0 M

Volume of CaF2 = 750 ml = 0.750 L

Molarity of CaF2 = 1.0 M

<u>To determine:</u>

The mass of CaCO3 produced

<u>Explanation:</u>

Na2CO3 + CaF2 → CaCO3 + 2NaF

Based on the reaction stoichiometry:

1 mole of Na2CO3 reacts with 1 moles of Caf2 to produce 1 mole of caco3

Moles of Na2CO3 present = V * M = 0.250 L * 6.0 moles/L = 1.5 moles

Moles of CaF2 present = V* M = 0.750 * 1 = 0.750 moles

CaF2 is the limiting reagent

Thus, # moles of CaCO3 produced = 0.750 moles

Molar mass of CaCO3 = 100 g/mol

Mass of CaCO3 produced = 0.750 moles * 100 g/mol  = 75 g

Ans: Mass of CaCO3 produced = 75 g

7 0
2 years ago
PLEASE HELP!!!!!
Doss [256]
A) an acid because it turns the paper blue
6 0
2 years ago
If a gas has a volume of 750 mL at 25oC, what would the volume of the gas be at 55oC?
blagie [28]

Answer:

The answer to your question is     V2 = 825.5 ml

Explanation:

Data

Volume 1 = 750 ml

Temperature 1 = 25°C

Volume 2= ?

Temperature 2 = 55°C

Process

Use the Charles' law to solve this problem

                V1/T1 = V2/T2

-Solve for V2

                V2 = V1T2 / T1

-Convert temperature to °K

T1 = 25 + 273 = 298°K

T2 = 55 + 273 = 328°K

-Substitution

                V2 = (750 x 328) / 298

-Simplification

                V2 = 246000 / 298

-Result

                V2 = 825.5 ml

             

7 0
2 years ago
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