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lutik1710 [3]
3 years ago
12

The pressure in a car tire is 198 kPa at 27°C. After a long drive, the pressure

Chemistry
1 answer:
AlekseyPX3 years ago
5 0

Answer : The temperature of the air in the tire is, 341 K

Explanation :

Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

P\propto T

or,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

where,

P_1 = initial pressure = 198 kPa

P_2 = final pressure = 225 kPa

T_1 = initial temperature = 27^oC=273+27=300K

T_2 = final temperature = ?

Now put all the given values in the above equation, we get:

\frac{198kPa}{300K}=\frac{225kPa}{T_2}

T_2=340.9K\approx 341K

Therefore, the temperature of the air in the tire is, 341 K

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The net ionic equation for the precipitation reaction that occurs when aqueous solutions of silver(i) nitrate and chromium(iii)
eduard

The net ionic equation formed is

       Ag^+(aq)+Cl^−(aq)→AgCl(s)

Chromium(III) nitrate and silver(I) chloride are the products of the balanced molecular equation for the reaction between chromium(III) chloride and silver(I) nitrate. An (s) next to the chemical formula for silver(I) chloride designates it as an insoluble salt.

      CrCl3(aq)+3AgNO3(aq)→Cr(NO3)3(aq)+3AgCl(s)

Silver and the chloride ions are the two ions that must interact to create silver(I) chloride. By designating ions as the reactants and silver(I) chloride as the product, the net ionic equation is formed.

      Ag^+(aq)+Cl^−(aq)→AgCl(s)

Ionic Equation:

In general, anions and cations react to generate a compound in a dissolved media, which is known as an ionic reaction. Water-insoluble salts are created when the ions of water-soluble salts interact with one another in an aqueous media.

To learn more about Ionic equaion click the given link

brainly.com/question/19705645

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2 years ago
For the Zn - Cu^2+ voltaic cell Zn(s) + Cu^2+(aq, 1M) + Cu(s) E degree _cell = 1.10 V Given that the standard reduction potentia
Fittoniya [83]

Answer : The value of E^o_{(Cu^{2+}/Cu)} is, 0.34 V

Explanation :

Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction:  Zn\rightarrow Zn^{2+}+2e^-

Reduction half reaction:  Cu^{2+}+2e^-\rightarrow Cu

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu

To calculate the E^o_{(Cu^{2+}/Cu)} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=E^o_{(Cu^{2+}/Cu)}-E^o_{(Zn^{2+}/Zn)}

Putting values in above equation, we get:

1.10V=E^o_{(Cu^{2+}/Cu)}-(-0.76V)

E^o_{(Cu^{2+}/Cu)}=0.34V

Hence, the value of E^o_{(Cu^{2+}/Cu)} is, 0.34 V

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Answer:

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