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Alecsey [184]
3 years ago
15

A force of 17 N is exerted on an area of 0.20 m2. What is the pressure?

Chemistry
2 answers:
miss Akunina [59]3 years ago
7 0
The required formula is

Pressure =  \frac{force}{area} \\\\Putting\ given\ values,\ we\ get\\\\Pressure =  \frac{17}{0.2} = 85\ Answer
solmaris [256]3 years ago
5 0
Pressure=\frac{Force}{Area}\\\\Force=17N\\\\Area=0,2m^2\\\\Pressure=\frac{17N}{0,2m^2}=85\ Pascals
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Explanation:

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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:

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(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}

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                               = -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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