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3 years ago

Balance the following chemical equation, then answer the following question.

1 answer:

3 0

Molar mass:

O2 = 31.99 g/mol
C8H18 = 144.22 g/mol

<span>2 C8H18(g) + 25 O2(g) = 16 CO2(g) + 18 H2O(g)

2 x 144.22 g --------------- 25 x 31.99 g
10.0 g ----------------------?? ( mass of O2)

10.0 x 25 x 31.99 / 2 x 144.22 =

7997.5 / 288.44 => 27.72 g of O2

hope this helps!

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The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas.

enot [183]

Answer:

$\large \boxed{\text{21.6 L}}$

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 180.16

C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g: 24.5

(a) Moles of C₆H₁₂O₆

$\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{24.5 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1360 mol C$_{6}$H$_{12}$O}_{6}$

(b) Moles of CO₂

$\text{Moles of CO}_{2} =\text{0.1360 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol CO}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.8159 mol CO}_{2}$

(c) Volume of CO₂

We can use the Ideal Gas Law.

pV = nRT

Data:

p = 0.960 atm

n = 0.8159 mol

T = 37 °C

(i) Convert the temperature to kelvins

T = (37 + 273.15) K= 310.15 K

(ii) Calculate the volume

$\begin{array}{rcl}pV &=& nRT\\\text{0.960 atm} \times V & = & \text{0.8159 mol} \times \text{0.082 06 L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{310.15 K}\\0.960V & = & \text{20.77 L}\\V & = & \textbf{21.6 L} \\\end{array}\\\text{The volume of carbon dioxide is $\large \boxed{\textbf{21.6 L}}$}$

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2 years ago

What determines the type of crystals in igneous rocks?

nika2105 [10]

Over all the size of the crystal depends on how fast magma will harden over time if the magma cools slowly the crystal will be larger than they would if the magma cooled faster

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3 years ago

How many moles of chlorine gas would occupy a volume of 35.5 L at a pressure of 100.0 kPa and a temperature of 100.0 degrees Cel

o-na [289]

1.137448506 mol moles of chlorine gas would occupy a volume of 35.5 L at a pressure of 100.0 kPa and a temperature of 100.0 degrees Celsius.

<h3>What is an ideal gas equation?</h3>

The ideal gas equation, pV = nRT, is an equation used to calculate either the pressure, volume, temperature or number of moles of a gas. The terms are: p = pressure, in pascals (Pa). V = volume, in $m^3$ .

We apply the formula of the ideal gases, we clear n (number of moles); we use the ideal gas constant R = 0.082 l atm / K mol:

PV= nRT

Given data:

P=100.0 kPa =0.986923 atm

T=100 degree celcius= 100 + 273 =373 K

V=35.5 L

Substituting the values in the equation.

n= $\frac{\;0,98 \;atm \;X \;35,5 \;L }{\;0,082\;atm / \;K mol \;X \;373 K}$

n= 1.137448506 mol

Hence, 1.137448506 mol moles of chlorine gas would occupy a volume of 35.5 L at a pressure of 100.0 kPa and a temperature of 100.0 degrees Celsius.

Learn more about ideal gas here:

brainly.com/question/16552394

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2 years ago

As much sugar as will dissolve is added to hot water. The water is then

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The answer is C because it says describe the final sugar and C is unsaturated and in the problem it says as much as sugar will dissovle .

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3 years ago

If Jerome is swinging on a rope and transferring energy from gravitational potential energy to kinetic energy, _____... is being

Pachacha [2.7K]

<u>Answer;</u>

-Work

If Jerome is swinging on a rope and transferring energy from gravitational potential energy to kinetic energy, <u>work </u> is being done.

<u>Explanation;</u>

Work refers to the application of a given force over a distance. Thus, we can say work is the product of force and distance.
Energy on the other hand is the ability of a body to change, its location, shape, or state of another body.
According to the work-Energy principle, a change in the kinetic energy, which is the energy possessed by a body in motion, is equivalent to the net work done on the body.

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3 years ago

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