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statuscvo [17]
3 years ago
6

Given the following values for the heats of formation, what is the number of moles of ethane (C2H6, MW 30.0) required to produce

1.00x10^3 kJ of heat by combustion to CO2 (g) and H2O (l).
C2H6 (g) + 7/2O2(g) → 2CO2 (g) + 3 H2O (l)

C2H6 (g) ΔHf° = -84.7 kJ/mol
CO2 (g) ΔHf° = -393.5 kJ/mol
H2O (l) ΔHf° = -285.8 kJ/mol

a. 0.595
b. 1.56
c. 0.641
d. can't be determined without ΔHfo for O2
e. 1.68
Chemistry
1 answer:
baherus [9]3 years ago
7 0

Answer:

0.641 moles of ethane

Explanation:

Based on the equation:

C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H2O(l)

We can determine ΔH of reaction using Hess's law. For this equation:

<em>Hess's law: ΔH products - ΔH reactants</em>

ΔH = {2ΔHCO2 + 3ΔHH2O} - {ΔHC2H6}

<em>Pure monoatomic substances have a ΔH = 0kJ/mol; ΔHO2 = 0kJ/mol</em>

<em />

ΔH = {2*-393.5kJ/mol + 3*-285.8kJ/mol} - {-84.7kJ/mol}

ΔH = -1559.7kJ/mol

That means when 1 mole of ethane is in combustion there are released 1559.7kJ of heat. To produce 1.00x10³kJ there are needed:

1.00x10³kJ * (1mole ethane / 1559.7kJ) =

<h3>0.641 moles of ethane</h3>
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Calculate the number of moles of solute in 27.55 mL of 0.1185 M K2Cr2O7(aq).
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Answer:

0.01185M = moles/0.02755L

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Explanation:

7 0
3 years ago
. A sample of crude oil has a density of 0.87 g/mL. What volume (in liters) does a 3.6 kg sample of this oil occupy
gayaneshka [121]

Answer:

The volume is 4.13793 L

Explanation:

Density is a quantity that expresses the relationship between the mass and the volume of a body, so it is defined as the quotient between the mass and the volume of a body:

density=\frac{mass}{volume}

Density is a characteristic property of every body or substance.

The most commonly used units of density are \frac{kg}{m^{3} } or \frac{g}{cm^{3} } for solids, and \frac{kg}{L} or \frac{g}{mL} for liquids and gases.

In this case, you know:

  • density= 0.87 \frac{g}{mL}
  • mass= 3.6 kg= 3,600 g (being 1 kg=1,000 g)
  • volume= ?

Replacing:

0.87\frac{g}{mL} =\frac{3,600 g}{volume}

Solving:

volume =\frac{3,600 g}{0.87\frac{g}{mL}}

volume= 4,137.93 mL

Being 1,000 mL=1 L, then volume= 4,137.93 mL= 4.13793 L

<u><em>The volume is 4.13793 L</em></u>

5 0
4 years ago
A 12 gram piece of metal is heated to 300 °C from 100 °C with 1120 Joules of energy. What is the specific heat of the metal?
Sergeeva-Olga [200]

Answer:

The specific heat for the metal is 0.466 J/g°C.

Explanation:

Given,

Q = 1120 Joules

mass = 12 grams

T₁ = 100°C

T₂ = 300°C

The specific heat for the metal can be calculated by using the formula

Q = (mass) (ΔT) (Cp)

ΔT = T₂ - T₁ = 300°C  - 100°C   = 200°C

Substituting values,

1120 = (12)(200)(Cp)

Cp = 0.466 J/g°C.

Therefore, specific heat of the metal is 0.466 J/g°C.

7 0
3 years ago
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Brilliant_brown [7]
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2.  A Bunsen burner was used to boil away the water from the salt water solution leaving only salt.

I hope this helps.  Let me know if anything is unclear.
4 0
3 years ago
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