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Sonbull [250]
2 years ago
7

Calculate the work (w) and ΔEo, in kJ, at 298 K and 1 atm pressure, for the combustion of one mole of C4H10 (g). First write and

balance the equation. The products will be CO2 (g) and H2O (g)
The value of ΔHo for this reaction is -2658.3 kJ/mol
Chemistry
1 answer:
Paul [167]2 years ago
3 0

Answer:

Explanation:

2C₄H₁₀ + 13O₂ = 8 CO₂ + 10H₂O

Change in number of moles Δn = 18 - 15 = + 3 moles .

ΔHo = -2658.3 kJ/mol.

ΔHo = ΔEo+ Δn RT

Δn = 3

For one mole Δn = 1.5

ΔHo = ΔEo+ W

W = Δn RT

= 1.5 x 8.31 x 298

= 3714.5 J

= 3.7 kJ /mole

ΔHo = ΔEo+ W

ΔEo =  ΔHo -  W

= -2658.3 - 3.7  kJ

= - 2662 kJ .

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A mass of 5.4 grams of aluminum (al) reacts with an excess of copper (ii) chloride (cucl2) in solution, as shown below. 3cucl2 2
horrorfan [7]

Answer:

Mass of copper produced is 19.07g

Explanation:

Let's bring out the chemical equation;

3CuCl2 + 2Al → 2AlCl3 + 3Cu

3           :   2       :   2       : 3

Upon confirming that the reaction is indeed balanced, we can proceed.

The questions asks to calculate mass of Cu formed when a mass of 5.4g of Al is being used.

From the equation, what is the relationship between Al and Cu?

2 mol of Al would react to form 3 mol of Cu

Expressing this in terms of mass, we have;

mass = no. of moles * molar mass

Mass of Aluminium = 2 * 26.98 = 53.98 g

Mass of Copper = 3 * 63.546 = 190.638g

This means;

53.98 g of Al would react to form 190.638g of Cu

So how much Cu would form from 5.4 g of Al?

This leads us to;

53.98 = 190.638

5.4 = x

Upon cross multiplication, we are left with;

x = (190.638 * 5.4) / 53.98

x = 19.07 g

Mass of copper produced is 19.07g

8 0
2 years ago
Hydrochloric acid (75.0 mL of 0.250 M) is added to 225.0 mL of 0.0550 M Ba(OH)2 solution. What is the concentration of the exces
Shalnov [3]

Answer:  The concentration of excess [OH^-] in solution is 0.017 M.

Explanation:

1. Molarity=\frac{moles}{\text {Volume in L}}

moles of HCl=Molarity\times {\text {Volume in L}}=0.250\times 0.075=0.019moles

1 mole of HCl give = 1 mole of H^+

Thus 0.019 moles of HCl give = 0.019 mole of H^+

2. moles of Ba(OH)_2=Molarity\times {\text {Volume in L}}=0.0550\times 0.225=0.012moles

According to stoichiometry:

1 mole of Ba(OH)_2 gives = 2 moles of OH^-

Thus 0.012 moles of Ba(OH)_2 give = 2 \times 0.012=0.024 moles of OH^-

H^++OH^-\rightarrow H_2O

As 1 mole of H^+ neutralize 1 mole of OH^-

0.019 mole of H^+ will neutralize 0.019 mole of OH^-

Thus (0.024-0.019)= 0.005 moles of OH^- will be left.

[OH^-]=\frac{\text {moles left}}{\text {Total volume in L}}=\frac{0.005}{0.3L}=0.017M

Thus molarity of [OH^-] in solution is 0.017 M.

4 0
2 years ago
Read 2 more answers
Suppose that on a hot and sticky afternoon in the spring, a tornado passes over the high school. If the air pressure in the lab
uranmaximum [27]

The answer is B) 230 m3

5 0
2 years ago
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A gold wire has a diameter of 1.00 mm. What length of this wire contains exactly 1.00 mol of gold? (density of Au = 17.0 g/cm3)
allochka39001 [22]

Answer:

The answer to your question is 7160 cm

Explanation:

Data

diameter = 1 mm

length = ?

amount of gold = 1 mol

density = 17 g/cm³

Process

1.- Get the atomic mass of gold

Atomic mass = 197 g

then, 197g ------------ 1 mol

2.- Calculate the volume of this wire

density = mass/volume

volume = mass/density

volume = 197/17

volume = 5.7 cm³

3.- Calculate the length of the wire

Volume = πr²h

solve for h

h = volume /πr²

radius = 0.05 cm

substitution

h = 5.7/(3.14 x 0.05²)

h = 5.7 / 0.0025

h = 7159.2 cm ≈ 7160 cm

8 0
2 years ago
Read 2 more answers
What is the final pressure (expressed in atm) of a 3.05 l system initially at 724 mm hg and 298 k, that is compressed to a final
krok68 [10]

<u>Answer:</u>

P2 = 778.05 mm Hg = 1.02 atm

<u>Explanation:</u>

We are to find the final pressure (expressed in atm)  of a 3.05 liter system initially at 724 mm hg and 298 K which is compressed to a final volume of 2.60 liter at 273 K.

For this, we would use the equation:

\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}

where P1 = 724 mm hg

V1 = 3.05 L

T1 = 298 K

P2 = ?

V2 = 2.6 L

T2 = 173 K

Substituting the given values in the equation to get:

\frac{(724)(3.05)}{298} =\frac{P_2(2.6)}{173}

P2 = 778.05 mm Hg = 1.02 atm

7 0
2 years ago
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