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Ivanshal [37]
3 years ago
14

The gases inside of a normal tire are usually measured when they are at ambient standard temperature (298 K) & under a contr

olled pressure of 32.5 PSI in order for the device to remain “stable” – not blow up! Sometimes, driving conditions cause the pressure inside the tire to reach 40.68 PSI, what is the temperature of the air inside the tire at 40.68 PSI (in °C)?
Chemistry
1 answer:
Anastaziya [24]3 years ago
8 0

Answer:

100°C

Explanation:

The following were obtained from the question:

Initial temperature (T1) = 298 K

Initial pressure (P1) = 32.5 Psi

Final pressure (P2) = 40.68 Psi

Final temperature (T2) =..?

The final temperature can be obtained as follow:

P1/T1 = P2/T2

32.5/298 = 40.68/T2

Cross multiply to express in linear form

32.5 x T2 = 298 x 40.68

Divide both side by 32.5

T2 = (298 x 40.68) /32.5

T2 = 373K

Next, we shall convert 373K to celsius.

This is illustrated below:

T(°C) = T(K) – 273

T(K) = 373

T(°C) = 373 – 273

T(°C) = 100°C

Therefore, the temperature of the air inside the tire at 40.68 Psi is 100°C

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A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 64.0 mL of 0.0600 M EDTA . Titration of the excess unreacted EDTA req
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Answer:

the concentration of Cd^{2+}  in the original solution= 0.0088 M

the concentration of Mn^{2+} in the original solution = 0.058 M

Explanation:

Given that:

The volume of the sample  containing Cd2+ and Mn2+ =  50.0 mL; &

was treated with 64.0 mL of 0.0600 M EDTA

Titration of the excess unreacted EDTA required 16.1 mL of 0.0310 M Ca2+

i.e the strength of the Ca2+ = 0.0310 M

Titration of the newly freed EDTA required 14.2 mL of 0.0310 M Ca2+

To determine the concentrations of Cd2+ and Mn2+ in the original solution; we have the following :

Volume of newly freed EDTA = \frac{Volume\ of \ Ca^{2+}* Sample \ of \ strength }{Strength \ of EDTA}

= \frac{14.2*0.0310}{0.0600}

= 7.3367 mL

concentration of  Cd^{2+} = \frac{volume \ of \  newly  \ freed \ EDTA * strength \ of \ EDTA }{volume \ of \ sample}

= \frac{7.3367*0.0600}{50}

= 0.0088 M

Thus the concentration of Cd^{2+} in the original solution = 0.0088 M

Volume of excess unreacted EDTA = \frac{volume \ of \ Ca^{2+} \ * strength \ of Ca^{2+} }{Strength \ of \ EDTA}

= \frac{16.1*0.0310}{0.0600}

= 8.318 mL

Volume of EDTA required for sample containing Cd^{2+}   and  Mn^{2+}  = (64.0 - 8.318) mL

= 55.682 mL

Volume of EDTA required for Mn^{2+}  = Volume of EDTA required for

                                                                sample containing  Cd^{2+}   and  

                                                             Mn^{2+} --  Volume of newly freed EDTA

Volume of EDTA required for Mn^{2+}  = 55.682 - 7.3367

= 48.3453 mL

Concentration  of Mn^{2+} = \frac{Volume \ of EDTA \ required \ for Mn^{2+} * strength \ of \ EDTA}{volume \ of \ sample}

Concentration  of Mn^{2+} =  \frac{48.3453*0.0600}{50}

Concentration  of Mn^{2+}  in the original solution=   0.058 M

Thus the concentration of Mn^{2+} = 0.058 M

6 0
3 years ago
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