Initial temperature of water (oC)

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:

The standard enthalpy of reaction = -4854.7kJ

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

Explanation:

The balanced chemical equation for the combustion of heptane:

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

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

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

Here, $\nu$ is the stoichiometric coefficient

⇒ $\Delta H _{r}^{\circ } =$

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

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

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

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

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

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

To calculate the difference: 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) = 9910.288 J = 9.91 kJ (∵ 1 kJ = 1000J )

8 0

3 years ago

Water exists in liquid form between its 1. Freezing and Boiling point 2. Freezing and melting point 3. Boiling and melting point

Zigmanuir [339]

Answer:

Freezing and boiling point

Explanation:

A liquid form of any substance is an intermediate form between the solid form and the gaseous form.

Decreasing the temperature of liquid water according to the phase diagram of $H_2O$ would freeze it and we would have a phase change from liquid to solid (ice) at the freezing point of water.

Similarly, heating water to its boiling point would evaporate water and we would have a phase change from liquid to gas (water vapor).

Therefore, liquid water exists between its freezing and boiling point.

8 0

3 years ago

So I am lost and I don't know how what any of those conversion factor things are and I'm wondering how to do #2,3, and 4. (I did

dalvyx [7]

Explanation:

#2.

A centigram is 1/100 of a gram, so that means a gram equals 100 centigrams.

Therefore you multiply 72.4 grams by 100/1 (or just 100), and get 7240 cg.

You did that one right but put the wrong unit in the answer. It is is cg ( centigrams).

#3.

1 liter is equal to 1000 milliliters, and I kiloliter is equal to 1000 liters. So one kiloliter is 1000*1000 milliliters or 1,000,000 milliliters.

The conversion factor would be

1/1000000

#4.

1 gigabyte is equal to 10^9 bytes.

I byte is equal to 10^9 bytes.

So 1 gigabyte is 10^9 * 10^9 nanobytes, or 10^18.

The conversion factor would be (1*10^18)/1.

6 0

3 years ago

How many atoms total are present in one molecule of 5-methyl-2 hexanol C7H16O

muminat

Answer:

24 atoms

Explanation:

The formula of the compound is C₇H₁₆O

The number of atoms that makes up one molecule of this compound are:

7 carbon atoms

16 hydrogen atoms

1 oxygen atom

The total number of atoms = 7+16+1 = 24 atoms

3 0

3 years ago

6.626 × 10⁻³⁴ × 7.08× 10¹⁴

PolarNik [594]

The answer is 4.69 x 10⁻¹⁹ I hope this helped!

3 0

3 years ago