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ohaa [14]
2 years ago
9

The mole fraction of oxygen molecules in dry air is 0.2095. What volume of dry air at 1.00 atm and 25°C is required for burning

completely 1.00 L of octane (C8H18, density = 0.7025 g/mL), yielding carbon dioxide and water?
Chemistry
1 answer:
Temka [501]2 years ago
0 0

Answer:

V = 8963 L

Explanation:

Our strategy here is to realize this problem involves stoichiometric calculation based on the balanced chemical equation for the combustion of octane:

C₈H₁₈ + 25/2 O₂ ⇒ 8 CO₂ + 9 H₂O

From the density of octane we can obtain the number of moles:

D = m/V ⇒ m= D x V = 0.7025 g/mL x ( 1000 mL) = 702.5 g

MW octane  = 702.5 g/ 114.23 g/mol = 6.15 mol

Required  mol oxygen to react with octane:

6.15 mol octane x  25/2 mol O₂ / mol octane = 76.8 mol O₂

Now mol fraction is given by mol O₂ / total number mol air ⇒

n air = 76.8 mol O₂ / (0.2095 mol O₂ / mol air )  = 366.59 mol air

and from the ideal gas law we can compute the volume of air:

PV = nRT  ⇒ V = nRT/P

V = 366.59 mol air x 0.08205 Latm/Kmol x (25+ 273) K/ 1 atm

   = 8,963 Lts

Note we treat here  air as a compund which is allowed in combustion problems.

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Assuming that the bath contains 250.0 g of water and that the calorimeter itself absorbs a negligible amount of heat, calculate
Anvisha [2.4K]

Answer:

-3272     kJ/mol

Explanation:

Given and known facts

Mass of Benzene = 0.187 grams

Mass of water = 250 grams

Standard heat capacity of water = 4.18 J/g∙°C

Change in temperature ΔT = 7.48°C

Heat

=250 * 4.18 * 7.48\\=7816.6 \\=7.82

Heat released by benzine is - 7.82 kJ

Now, we know that

0.187 grams of benzene release = -7.82  kJ heat

So, 1 g benzine releases

\frac{ -7.82 }{0.187}\\= -41.8

kJ/g

0.187 * \frac{1}{78.108}=0.00239 mol C6H6

Heat released

= \frac{-7.82}{ 0.00239}

=-3272     kJ/mol

4 0
2 years ago
What’s the bladder wall made up of?
AleksandrR [38]

Answer:

The bladder wall is made of many layers, including: Urothelium or transitional epithelium. This is the layer of cells that lines the inside of the kidneys, ureters, bladder, and urethra. Cells in this layer are called urothelial cells or transitional cells

Explanation:

8 0
3 years ago
1. BaBr2(aq) + H2SO4(aq) →?<br> please balance the equation and predict the products
Andrew [12]

Answer:

BaBr2 (aq) + H2SO4 (aq) → BaSO4 (s) + 2 HBr (aq)

Explanation:

This is a precipitation reaction: BaSO4 is the formed precipitate.

4 0
3 years ago
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8 0
3 years ago
An electron in the hydrogen atom makes a transition from an energy state of principal quantum number ni to the n = 2 state. If t
topjm [15]

Answer:

\boxed{3}

Explanation:

The Rydberg equation gives the wavelength λ for the transitions:

\dfrac{1}{\lambda} = R \left ( \dfrac{1}{n_{i}^{2}} - \dfrac{1}{n_{f}^{2}} \right )

where

R= the Rydberg constant (1.0974 ×10⁷ m⁻¹) and

\text{$n_{i}$ and $n_{f}$ are the numbers of the energy levels}

Data:

n_{f} = 2

λ = 657 nm

Calculation:  

\begin{array}{rcl}\dfrac{1}{657 \times 10^{-9}} & = & 1.0974 \times 10^{7}\left ( \dfrac{1}{2^{2}} - \dfrac{1}{n_{f}^{2}} \right )\\\\1.522 \times 10^{6} &= &1.0974\times10^{7}\left(\dfrac{1}{4} - \dfrac{1}{n_{f}^{2}} \right )\\\\0.1387 & = &\dfrac{1}{4} - \dfrac{1}{n_{f}^{2}} \\\\-0.1113 & = & -\dfrac{1}{n_{f}^{2}} \\\\n_{f}^{2} & = & \dfrac{1}{0.1113}\\\\n_{f}^{2} & = & 8.98\\n_{f} & = & 2.997 \approx \mathbf{3}\\\end{array}\\\text{The value of $n_{i}$ is }\boxed{\mathbf{3}}

7 0
2 years ago
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