Use the periodic to fill in the numbers in the electron configurations shown below

Consider the following reaction at constant P. Use the information here to determine the value of Î”Ssurr at 355 K. Predict whet

Sloan [31]

The given reaction is as follows:

2NO (g) + O₂ (g) = 2NO₂ (g), ΔH = -114 kJ

It is known that dSsurr = -dHsys / T (Temp = 355 K)

So, dSsurr = - (-114 × 1000) / 355

dSsurr = +321.12 J/K

Hence, the value of dSsurr is +321.12 J/K

For a reaction to be spontaneous, dG<0,

Also dStotal = dSsys + dSsurr > 0

It is known that dG = dHsys - TdSsys,

Now let us assume,

dG<0

Also, dStotal = dSsys + dSsurr > 0

(-114 × 1000) - (355 × dSsys) <0

355 × dSsys > -114 × 1000

dSsys > -321

dSsys >dSsurr

dSsys + dSsurr > 0

dStotal > 0

Thus, the assumption is correct, and the given reaction is spontaneous. Hence, the final answer is Ssurr = +321 J/K reaction is spontaneous.

8 0

2 years ago

Which of the following is used to show sound waves?

Ipatiy [6.2K]

I think it’s c bc it makes more sense

3 0

2 years ago

Consider the reaction 2 SO2(g) + O2(g) <=> 2 SO3(g), which is exothermic as written. What would be the effect on the equil

enot [183]

Answer:

Removing O₂, means removing one of the reactants and the system would counteract this effect by producing more O₂, thereby shifting the equilibrium position to the left and favouring the backward reaction.

Explanation:

The principle that explains how changes in temperature, Concentration and Pressure of reactants or products of a reaction at equilibrium affect the equilibrium position of the reaction is the Le Chatelier's principle.

The Principle explains that a system/process if a system/process which is at equilibrium is disturbed/perturbed/constrained by one or more changes (in concentration, pressure or temperature), the system would shift the equilibrium position to counteract the effects of this change.

Removing O₂, means removing one of the reactants (changing its concentration) and the system would counteract this effect by producing more O₂, thereby shifting the equilibrium position to the left and favouring the backward reaction.

5 0

2 years ago

Two reasons that a plant from a rain forest would not survive in a desert home?

ss7ja [257]

Answer: A planet from a rain forest would not survive in a desert home, for the fact they depend on water and as these live from water, without it, this leads to dehydration and with loss of water with plants, nothing is able to survive when it comes to a rain forest plant being in a desert home. Another reason is the high trees and leaves that are providing the plants enough sunlight or shade to grow, it guarantees to help keep the temperature normal. Being in a deserted area would mean that there would be a temperature change, something the plant is not used to. Without that needed shade and avoiding the scorching sun, they will die out from how different the temperature is, and how hot it is.

Explanation: I hope this helped you.

4 0

2 years ago

Read 2 more answers

The two isotopes of chlorine are LaTeX: \begin{matrix}35\\17\end{matrix}Cl35 17 C l and LaTeX: \begin{matrix}37\\17\end{matrix}C

Kay [80]

Answer: The percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 77.5% and 22.5% respectively.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope be 'x'. So, fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope will be '1 - x'

For $_{17}^{35}\textrm{Cl}$ isotope:

Mass of $_{17}^{35}\textrm{Cl}$ isotope = 35 amu

Fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope = x

For $_{17}^{37}\textrm{Cl}$ isotope:

Mass of $_{17}^{37}\textrm{Cl}$ isotope = 37 amu

Fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope = 1 - x

Average atomic mass of chlorine = 35.45 amu

Putting values in equation 1, we get:

$35.45=[(35\times x)+(37\times (1-x))]\\\\x=0.775$

Percentage abundance of $_{17}^{35}\textrm{Cl}$ isotope = $0.775\times 100=77.5\%$

Percentage abundance of $_{17}^{37}\textrm{Cl}$ isotope = $(1-0.775)=0.225\times 100=22.5\%$

Hence, the percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 77.5% and 22.5% respectively.

6 0

2 years ago