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pochemuha
2 years ago
11

Carla makes a small cardboard boat and floats it in a tub of water. When she adds a drop of dish soap behind the boat, the boat

moves forward. Which statement best explains what happened? The soap increased the viscosity of the water behind the boat, allowing the water in front of the boat to flow more easily. The soap reduced the surface tension behind the boat, causing both water molecules and the boat to be pulled forward. The soap increased the kinetic energy of the water molecules behind the boat, causing both the molecules and the boat to move forward. The soap formed an immiscible layer on the water, causing the soap to spread out and push the boat forward.
Chemistry
2 answers:
Mashcka [7]2 years ago
4 0

Answer:

The soap reduced the surface tension behind the boat, causing both water molecules and the boat to be pulled forward.

Explanation:

big brain/ im built different

butalik [34]2 years ago
3 0

Answer:

B

Explanation:

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How many meters are present in the 12.45 miles? Please show your work, and report your answer with the correct number of signifi
MatroZZZ [7]

Explanation:

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3 0
2 years ago
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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