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Svetradugi [14.3K]
3 years ago
14

If anyone is good at chemistry do you mind helping :)

Chemistry
1 answer:
Dominik [7]3 years ago
5 0

Answer:

• 1.62432 moles of nitrogen

• Tire Pressure: 2.74 * 10⁵ Pa

• The tires will burst

• Pressure: 244 kPa

Explanation:

• We can determine the number of moles of nitrogen using the formula pV = nRT, where p = pressure, V = volume, n = number of moles, R = gas constant, and T = absolute temperature.

Now remember we have our initial pressure in kilopascals so let's convert to pascals (249 pascals). The volume is given in liters, so let's convert into m². And the initial temperature is given in Celsius ⇒ our absolute temperature in Kelvins.

\mathrm{p\:}=\mathrm{249 kPa\:} = \mathrm{2.49 * 10^5\:},\\\mathrm{15.6L\:} =\mathrm{0.0156m^2\:},\\\mathrm{R\:}=\mathrm{8.314J/mol*K\:},\\\mathrm{T\:}=\mathrm{21C\:} + \mathrm{273\:}=\mathrm{294K\:}

Respectively the moles of nitrogen in each tire should be:

\mathrm{n\:}=\mathrm{pV/RT\:}=\mathrm{(2.49*10^5)(0.0156)/(8.314)(294)\:}=\frac{\left(2.49\cdot \:10^5\right)\left(0.0156\right)}{\left(8.134\right)\left(294\right)}=\frac{3884.4}{2391.396}\\

= 1.62432\dots \mathrm{moles\:}\mathrm{of\:}\mathrm{nitrogen\:}

• We can solve this part similarly. All our values will be the same, besides the temperature, as we have to consider both the initial and final temperature here.

\mathrm{T_2\:}=\mathrm{51C+ 273\:} }=\mathrm{324K\:} } -

\mathrm{p_2\:}=\mathrm{(2.49*10^5)(324)/(294)\:} }=\frac{\left(2.49\cdot \:10^5\right)\left(324\right)}{294}=\frac{40338000}{147}=274408.16326\dots

=2.74408.16326*10^5\dots\mathrm{Pa}

• The text mentions that the tires will burst when the internal pressure reaches 269kP. From part #2 we know that the final pressure will be, in kilopascals, 274kP. As 274 > 269, the tires will burst in Death Valley.

• We would want the final temperature = breaking pressure. Therefore,

\mathrm{p_2\:}=\mathrm{(269)(294)/(324)\:} }=\frac{79086}{324}=\frac{13181}{54}=244.09259\dots\mathrm{kPa\:} }

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Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H^o

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\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]

For the given chemical reaction:

2CO_2(g)+5H_2(g)\rightarrow C_2H_2(g)+4H_2O(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(C_2H_2(g))})+(4\times \Delta H^o_f_{(H_2O(g))})]-[(2\times \Delta H^o_f_{(CO_2(g))})+(5\times \Delta H^o_f_{(H_2(g))})]

We are given:

\Delta H^o_f_{(H_2O(g))}=-241.8kJ/mol\\\Delta H^o_f_{(H_2(g))}=0kJ/mol\\\Delta H^o_f_{(C_2H_2(g))}=226.7kJ/mol\\\Delta H^o_{rxn}=81.1kJ

Putting values in above equation, we get:

81.1=[(1\times (226.7)})+(4\times (-241.8))]-[(2\times \Delta H^o_f_{(CO_2(g))})+(5\times (0))]\\\\\Delta H^o_f_{(CO_2(g))}=-410.8kJ/mol

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(b) ₁₀Ne: 1s² 2s² 2p⁶

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Let the 2nd isotope be B

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2nd isotope (B):

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Atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100] + [(Mass of C × C%)/100]

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