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bixtya [17]
3 years ago
8

g When considering the effects of temperature on spontaneity, if both ΔH and ΔS are positive, _______. Select the correct answer

below: the process is spontaneous at all temperatures
Chemistry
1 answer:
Margaret [11]3 years ago
6 0

Explanation:

The spontaneity of a system is deduced by the sign of the gibbs free energy value. If it is negative, it means the process / reaction is spontaneous however a positive value indicates the such process is not spontaneous.

Gibbs free energy, enthalpy and entropy are related by the following equation;

ΔG = ΔH - TΔS

A positive value of enthalpy, H and entropy, S means that G would always be a negative value at all temperatures.

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It is representing syncline rock formation

there are two rock formation based on fold formation : syncline and anticline

In syncline rock formation there is fold like trough unlike anticline where it is like crust

In syncline the fold is downward as shown in photo and the new rock is outer fold and old at inner side

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Distance divided by time
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mote1985 [20]
The correct answer ; coal
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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 $\large \boxed{\textbf{21.6 L}}$}

7 0
3 years ago
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