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2 years ago

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A fuel was burned for 5 min, increasing the temperature of 10.0 g of water with a density of 1.00 g/ml by 9.0 oC. The fuel relea

sed _____ ca

1 answer:

jekas [21]2 years ago

7 0

The fuel released 90 calories of heat.

Let suppose that water experiments an entirely <em>sensible</em> heating. Hence, the heat released by the fuel is equal to the heat <em>absorbed</em> by the water because of principle of energy conservation. The heat <em>released</em> by the fuel is expressed by the following formula:

$Q = m\cdot c \cdot \Delta T$ (1)

Where:

$m$ - Mass of the sample, in grams.
$c$ - Specific heat of water, in calories per gram-degree Celsius.
$\Delta T$ - Temperature change, in degrees Celsius.

If we know that $m = 10\,g$ , $c = 1\,\frac{cal}{g\cdot ^{\circ}C}$ and $\Delta T = 9\,^{\circ}C$ , then the heat released by the fuel is:

$Q = (10\,g)\cdot \left(1\,\frac{cal}{g\cdot ^{\circ}C} \right)\cdot (9\,^{\circ}C)$

The fuel released 90 calories of heat.

We kindly invite to check this question on sensible heat: brainly.com/question/11325154

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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 )

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Note down the formula below

$\boxed{\sf Mass\%\;of\; element=\dfrac{Mass\:of\:the\: element}{Mass\;of\:the\: compound}\times 100}$

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$\\ \sf\longmapsto \dfrac{15}{135}\times 100$

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