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Marrrta [24]
3 years ago
7

.92 moles of carbon dioxide, CO2, was produced when you burned C6H120. (glucose) with O2.

Chemistry
1 answer:
Radda [10]3 years ago
6 0

Answer:

16.56g.

Explanation:

You need to compare the number of coefficient of the reaction product to find how much water produced. The reaction formula will produce 6 carbon dioxide(CO2) and 6 water (H2O).

If the reaction produces 0.92 moles of carbon dioxide, then the amount of water produced in moles will be: 0.92 moles * (6/6)= 0.92 moles

The molar mass of water is 18g/mol, so the calculation of moles to mass will be: 0.92 moles * (18g/mol)= 16.56g.

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Write a balanced equation for the combustion of C7H16(l) (heptane) -- i.e. its reaction with O2(g) forming the products CO2(g) a
JulsSmile [24]

Answer:

<u>The standard enthalpy of reaction = -4854.7kJ</u>

<u>The difference: </u>ΔH-ΔE = Δ(PV) = Δn.R.T = <u>9910.288 J ≈ 9.91 kJ</u>    

Explanation:

<u>The balanced chemical equation for the combustion of heptane</u>:

C₇H₁₆ (l) + 11 O₂ (g) → 7 CO₂ (g) + 8 H₂O (l)

Given: The standard enthalpy of formation (\Delta H _{f}^{\circ }) for: C₇H₁₆ (l) = -187.8 kJ/mol, O₂ (g) = 0 kJ/mol, CO₂ (g) = -393.5 kJ/mol, H₂O (l) = -286 kJ/mol

<u>To calculate the standard enthalpy of reaction (\Delta H _{r}^{\circ }) can be calculated by the Hess's law</u>:

\Delta H _{r}^{\circ } = \left [\sum \nu \cdot\Delta H _{f}^{\circ }(products)  \right ] - \left [\sum \nu\cdot\Delta H _{f}^{\circ }(reactants)  \right ]

Here, \nu is the stoichiometric coefficient

⇒ \Delta H _{r}^{\circ } =

\left [ 7\times \Delta H _{f}^{\circ }\left (CO_{2}\right )+ 8\times \Delta H _{f}^{\circ }\left (H_{2}O \right )\right ]

- \left [1\times \Delta H _{f}^{\circ }\left (C_{7}H_{16}\right ) +11\times \Delta H _{f}^{\circ }\left (O_{2} \right ) \right ]

=\left [ 7\times \left (-393.5 kJ/mol \right )+ 8\times \left (-286 kJ/mol \right )\right ]

-\left [1\times \left (-187.8 kJ/mol \right ) +11\times \left (0 kJ/mol \right ) \right ]

⇒ \Delta H _{r}^{\circ } = \left [ \left (-2754.5 \right )+ \left (-2288 \right )\right ]\left -[ \left (-187.8 \right ) +\left (0 \right )\right ]

⇒ \Delta H _{r}^{\circ } = \left [ -5042.5 ]\left -[ -187.8] = \left ( -4854.7kJ \right )

<u>To calculate the difference: </u>ΔH-ΔE=Δ(PV)

We use the ideal gas equation: P.V = n.R.T

⇒ ΔH-ΔE=Δ(PV) = Δn.R.T

Given: Temperature:T = 298K, R = 8.314 J⋅K⁻¹⋅mol⁻¹

Δn = number of moles of gaseous products - number of moles of gaseous reactants = (7)- (11) = (-4)

⇒ ΔH-ΔE=Δ(PV) = Δn.R.T = (-4 mol) × (8.314 J⋅K⁻¹⋅mol⁻¹) × (298K) = <u>9910.288 J = 9.91 kJ</u>                              (∵ 1 kJ = 1000J )

                                                                             

8 0
3 years ago
If you have a 1500 g aluminum pot, how much heat energy is needed to raise its temperature by 100°C?
Nataly [62]

The heat energy required to raise the temperature of 1500 g of aluminium pot by 100°C is 135 kJ.

The heat energy required to raise the temperature of 1500 g of copper pot by 100 °C is 57.75 kJ.

Explanation:

The heat energy required to raise the temperature of any body can be obtained from the specific heat formula. As this formula states that the heat energy required to raise the temperature of the body is directly proportional to the product of mass of the body, specific heat capacity of the material and temperature change experienced by the material.

So in this problem, the mass of the aluminium is given as m = 1500 g, the specific heat of the aluminium is 0.900 J/g °C. Then as it is stated that the temperature is raised by 100 °C, so the pots are heat to increase by 100 °C from its initial temperature. This means the difference in temperature will be 100°C (ΔT = 100°C).

Then, the heat energy required to raise the temperature will be

q = m*c*del T = 1500 * 0.900 * 100 = 135000 = 135 kJ

Thus, the heat energy required to raise the temperature of 1500 g of aluminium pot by 100 °C is 135 kJ.

Similarly, the mass of copper pot is given as 1500 g, the specific heat capacity of copper is 0.385 and the difference in temperature is 100  °C.

Then, the heat energy required to raise its temperature will be

q = m*c*del T = 1500 * 0.385 * 100 = 57750 = 57.75 kJ

And the heat energy required to raise the temperature of 1500 g of copper pot by 100°C is 57.75 kJ.

So, the heat energy required to raise the temperature of 1500 g of aluminium pot by 100°C is 135 kJ. And the heat energy required to raise the temperature of 1500 g of copper pot by 100 °C is 57.75 kJ.

8 0
3 years ago
What mass of Iron (III) acetate will be produced when 1.50 grams of iron reacts in acetic acid?
pantera1 [17]

Answer:

2Fe + 6HC2H3O2 → 2Fe(C2H3O2)3 + 3H2

Explanation:

There you go

4 0
2 years ago
Where are electrons received and donated at in a galvanic cell?
Vikki [24]

Explanation:

In a galvanic cell, the cathode is positively charged and the anode is negatively charged.

The cathode attracts electron while the anode donates or releases electrons.

Electrons received - Cathode

Electrons donated - Anode

3 0
3 years ago
According to the principle of superposition, which statement is true about the sediment layers shown here?Required to answer. Si
podryga [215]

if layer 4 is at the very bottom it would be c

3 0
2 years ago
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