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KiRa [710]
3 years ago
11

The combustion reaction of isopropyl alcohol is given below: C 3 H 7 O H ( l ) + 9 2 O 2 ( g ) → 3 C O 2 ( g ) + 4 H 2 O ( g ) T

he heat of combustion of isopropyl alcohol is -2006 kJ/mol. Given the following information, calculate the heat of formation of isopropyl alcohol.
Chemistry
1 answer:
jek_recluse [69]3 years ago
3 0

Answer:

the heat of formation of isopropyl alcohol is -317.82 kJ/mol

Explanation:

The heat of combustion of isopropyl alcohol is given as follows;

C₃H₇OH (l) +(9/2)O₂ → 3CO₂(g) + 4H₂O (g)

The heat of combustion of CO₂ and H₂O are given as follows

C (s) + O₂ (g) → CO₂(g) = −393.50 kJ

H₂ (g) + 1/2·O₂(g)   →  H₂O (l) = −285.83 kJ

Therefore we have

3CO₂(g) + 4H₂O (g) → C₃H₇OH (l) +(9/2)O₂ which we can write as

3C (s) + 3O₂ (g) → 3CO₂(g) = −393.50 kJ × 3  =

4H₂ (g) + 2·O₂(g)   →  4H₂O (l) = −285.83 kJ × 4

3CO₂(g) + 4H₂O (g) → C₃H₇OH (l) +(9/2)O₂ = +2006 kJ/mol

-1180.5 - 1143.32 +2006 = -317.82 kJ/mol

Therefore, the heat of formation of isopropyl alcohol = -317.82 kJ/mol.

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How many moles of KMnO, are in 66.38 g of KMnO?
schepotkina [342]

Answer:.603moles

Explanation:do given over 1 so 66.38 over 1 then multiply by 1 over 110.035344(the atomic mass of KMnO) and then you get the answear

5 0
3 years ago
How much energy is required to vaporize 0.5kg of water?
White raven [17]

Answer:

Explanation:

Vaporization Equation => Q = m·ΔHv

m = 0.5Kg water = 500g water

ΔHv = Heat of vaporization = 540 calories/gram

∴Q = 500g x 540cal/g = 270,000 calories x 4.184 joules/cal = 1,129,680 joules = 1130 Kilojoules (3 sig. figs.)

3 0
3 years ago
Convert 2.55 moles of water to grams.
Vlada [557]

Explanation:

No of mass = no of moles × molar mass

=2.55 × 18 g

= 45.9 g

6 0
3 years ago
An electrochemical cell has the following standard cell notation: Al(s) | Al3+(aq) || Mg2+(aq) | Mg(s)
Soloha48 [4]

Answer:

a. Al(s) ⇄ Al⁺³(aq) + 3e⁻ (oxidation)

Mg²⁺(aq) + 2e⁻ ⇄ Mg(s) (reduction)

b. ΔE° = + 0.715 V

c.  It's an electrolytic cell, because it's a nonspontaneous reaction.

d. 2Al(s) + 3Mg²⁺(aq) ⇄ 2Al⁺³(aq) + 3Mg(s)

Explanation:

a. By the notation given, first is represented the oxidation reaction and then the reduction reaction, so they are:

Al(s) ⇄ Al⁺³(aq) + 3e⁻ (oxidation)

Mg²⁺(aq) + 2e⁻ ⇄ Mg(s) (reduction)

b. The standard potential of the cell (ΔE°) is the reduction potential of the oxidation less the reduction potential of the reduction. The reduction potentials are:

Al(s) = -1.66 V

Mg(s) = -2.375 V

ΔE° = -1.66 - (-2.375)

ΔE° = + 0.715 V

c. It's an electrolytic cell.

A galvanic cell is spontaneous, so the cathode (reduction) has a higher E° than the cathode (oxidation). In this case, the oxidation reaction has a higher E°, so the reaction is nonspontaneous and it's necessary an external force to it happen, so it's an electrolytic cell.

d. 2Al(s) + 3Mg²⁺(aq) ⇄ 2Al⁺³(aq) + 3Mg(s)

The number of electrons must be the same, so the oxidation reaction is multiplied by 2, and the reduction reaction by 3.

5 0
3 years ago
A gas starts at a volume of 23 L and a pressure of 1.23 atm. What is the new pressure if you
Anna71 [15]

Answer:

<h2>1.89 atm</h2>

Explanation:

The new volume can be found by using the formula for Boyle's law which is

P_1V_1 = P_2V_2

Since we are finding the new volume

V_2 =  \frac{P_1V_1}{P_2}  \\

From the question we have

V_2 =  \frac{1.23 \times 23}{15}  =  \frac{28.29}{15}  \\  = 1.886

We have the final answer as

<h3>1.89 atm</h3>

Hope this helps you

7 0
3 years ago
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