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\:}](https://tex.z-dn.net/?f=%5Cmathrm%7Bp%5C%3A%7D%3D%5Cmathrm%7B249%20kPa%5C%3A%7D%20%3D%20%5Cmathrm%7B2.49%20%2A%2010%5E5%5C%3A%7D%2C%5C%5C%5Cmathrm%7B15.6L%5C%3A%7D%20%3D%5Cmathrm%7B0.0156m%5E2%5C%3A%7D%2C%5C%5C%5Cmathrm%7BR%5C%3A%7D%3D%5Cmathrm%7B8.314J%2Fmol%2AK%5C%3A%7D%2C%5C%5C%5Cmathrm%7BT%5C%3A%7D%3D%5Cmathrm%7B21C%5C%3A%7D%20%2B%20%5Cmathrm%7B273%5C%3A%7D%3D%5Cmathrm%7B294K%5C%3A%7D)
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}\\](https://tex.z-dn.net/?f=%5Cmathrm%7Bn%5C%3A%7D%3D%5Cmathrm%7BpV%2FRT%5C%3A%7D%3D%5Cmathrm%7B%282.49%2A10%5E5%29%280.0156%29%2F%288.314%29%28294%29%5C%3A%7D%3D%5Cfrac%7B%5Cleft%282.49%5Ccdot%20%5C%3A10%5E5%5Cright%29%5Cleft%280.0156%5Cright%29%7D%7B%5Cleft%288.134%5Cright%29%5Cleft%28294%5Cright%29%7D%3D%5Cfrac%7B3884.4%7D%7B2391.396%7D%5C%5C)
![= 1.62432\dots \mathrm{moles\:}\mathrm{of\:}\mathrm{nitrogen\:}](https://tex.z-dn.net/?f=%3D%201.62432%5Cdots%20%5Cmathrm%7Bmoles%5C%3A%7D%5Cmathrm%7Bof%5C%3A%7D%5Cmathrm%7Bnitrogen%5C%3A%7D)
• 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{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](https://tex.z-dn.net/?f=%5Cmathrm%7Bp_2%5C%3A%7D%3D%5Cmathrm%7B%282.49%2A10%5E5%29%28324%29%2F%28294%29%5C%3A%7D%20%7D%3D%5Cfrac%7B%5Cleft%282.49%5Ccdot%20%5C%3A10%5E5%5Cright%29%5Cleft%28324%5Cright%29%7D%7B294%7D%3D%5Cfrac%7B40338000%7D%7B147%7D%3D274408.16326%5Cdots)
![=2.74408.16326*10^5\dots\mathrm{Pa}](https://tex.z-dn.net/?f=%3D2.74408.16326%2A10%5E5%5Cdots%5Cmathrm%7BPa%7D)
• 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\:} }](https://tex.z-dn.net/?f=%5Cmathrm%7Bp_2%5C%3A%7D%3D%5Cmathrm%7B%28269%29%28294%29%2F%28324%29%5C%3A%7D%20%7D%3D%5Cfrac%7B79086%7D%7B324%7D%3D%5Cfrac%7B13181%7D%7B54%7D%3D244.09259%5Cdots%5Cmathrm%7BkPa%5C%3A%7D%20%7D)
Explanation:
Method of prepration of sodium thiosulphate - definition
In the laboratory, this salt can be prepared by heating an aqueous solution of sodium sulphite with sulphur or by boiling aqueous NaOH and sulfur according to this equation:
![6NaOH + 4 S _{2} Na _{2} S ->Na _{2} S_{2}O_{3}+ 3H _{2}O.](https://tex.z-dn.net/?f=6NaOH%20%2B%204%20S%20_%7B2%7D%20Na%20_%7B2%7D%20S%20-%3ENa%20_%7B2%7D%20S_%7B2%7DO_%7B3%7D%2B%203H%20_%7B2%7DO.)
You know since you wanna go around acting goofy...
i think you deserve a CERTIFIED CLOWN LICENSE
NaHCO3 = 22.99 + 1.008 + 16(3) = 83.99 g/mol
<span>Na = 22.99g/83.99 g weight of molecule =.2727 or 27.27% </span>
<span>3.0 g* .2727 = 0.8211 grams of sodium in sample of NaHCO3
</span><span>0.8211 grams Na + 1.266 grams Cl = 2.087 grams</span>