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AveGali [126]
3 years ago
5

Select all the correct answers.

Chemistry
2 answers:
swat323 years ago
8 0

Silver tarnishes and becomes black when exposed to oxygen:

              Diesel fuel burns when it is heated:

Mashutka [201]3 years ago
7 0

Answer:

              Silver tarnishes and becomes black when exposed to oxygen:

              Diesel fuel burns when it is heated:

Explanation:

                    Chemical Changes are those changes in which the chemistry at molecular level is changed as the starting material is converted in to a new and different final material. This change takes place with the breaking of old bonds and forming of new bonds respectively.

Silver tarnishes and becomes black when exposed to oxygen:

This is a chemical change as Silver a white and lustrous starting material is being converted into a tarnish black final material. However, Silver doesn't readily react with oxygen at normal conditions. It readily react with Sulfur containing compounds in air and produces black compound as Ag₂S.

Diesel fuel burns when it is heated:

Diesel is a mixture of hydrocarbons ranging approximately from C₁₀H₂₀ to C₁₅H₂₈. When these hydrocarbons are burnt they produces a new materials i.e. CO₂ and H₂O.

                             C₁₀H₂₀  +  15 O₂    →     10 CO₂  + 10 H₂O

Hence, it is also a chemical reaction.

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calculate the difference in slope of the chemical potential against temperature on either side of the normal freezing point of w
kipiarov [429]

Answer:

(a) The normal freezing point of water (J·K−1·mol−1) is -22Jmole^-^1k^-^1

(b) The normal boiling point of water (J·K−1·mol−1) is -109Jmole^-^1K^-^1

(c) the chemical potential of water supercooled to −5.0°C exceed that of ice at that temperature is  109J/mole

Explanation:

Lets calculate

(a) - General equation -

      (\frac{d\mu(\beta )}{dt})p-(\frac{d\mu(\alpha) }{dt})_p = -5_m(\beta )+5_m(\alpha ) =  -\frac{\Delta H}{T}

 \alpha ,\beta → phases

ΔH → enthalpy of transition

T → temperature transition

 (\frac{d\mu(l)}{dT})_p -(\frac{d\mu(s)}{dT})_p == -\frac{\Delta_fH}{T_f}

            = \frac{-6.008kJ/mole}{273.15K} ( \Delta_fH is the enthalpy of fusion of water)

           = -22Jmole^-^1k^-^1

(b) (\frac{d\mu(g)}{dT})_p-(\frac{d\mu(l)}{dT})_p= -\frac{\Delta_v_a_p_o_u_rH}{T_v_a_p_o_u_r}

                                  = \frac{40.656kJ/mole}{373.15K} (\Delta_v_a_p_o_u_rH is the enthalpy of vaporization)

                               = -109Jmole^-^1K^-^1

(c) \Delta\mu =\Delta\mu(l)-\Delta\mu(s) =-S_m\DeltaT

[\mu(l-5°C)-\mu(l,0°C)] =  [\mu(s-5°C)-\mu(s,0°C)]=-S_mΔT

\mu(l,-5°C)-\mu(s,-5°C)=-Sm\DeltaT [\mu(l,0

\Delta\mu=(21.995Jmole^-^1K^-^1)\times (-5K)

     = 109J/mole

6 0
2 years ago
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