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Andrei [34K]
3 years ago
13

The following characteristics describe which of these terms? The substance is mixed uniformly throughout and each part of the su

bstance contains the same ratio of materials with the same properties.
A.homogeneous mixture
B. compound
C. heterogeneous mixture
D. pure substance
Chemistry
2 answers:
katrin [286]3 years ago
8 0
Hi there! I had the same question on my test as well!
The correct answer is (A) Homogeneous mixture!
I hope I helped you, and I am 100% this is the right answer!
prohojiy [21]3 years ago
6 0

Answer: The correct answer is (A).

Explanation:

Homogeneous Mixtures are the mixture in which:

  • Substance is uniformly distributed throughout the medium.
  • Each part of the mixture contains same ratio of the substance.
  • Properties of the substance remains unchanged.
  • For example: Sugar in water, alloys of metals etc.

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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
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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

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Answer:

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Explanation:

5 0
3 years ago
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