The distance covered will be:

$d = v_0\cdot t + \frac{1}{2}a\cdot t^2 = 20\frac{m}{s}\cdot 5\ s + \frac{5}{2}\frac{m}{s^2}\cdot 5^2\ s^2 = \bf 162,5\ m$

The correct answer is D).

8 0

3 years ago

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The freezing-point depression of a 0.100 m MgSO4 solution is 0.225°C. Determine the experimental van't Hoff factor of MgSO4 at t

Andrews [41]

Answer: The experimental van't Hoff factor is 1.21

Explanation:

The expression for the depression in freezing point is given as:

$\Delta T_f=iK_f\times m$

where,

i = van't Hoff factor = ?

$\Delta T_f$ = depression in freezing point = 0.225°C

$K_f$ = Cryoscopic constant = 1.86°C/m

m = molality of the solution = 0.100 m

Putting values in above equation, we get:

$0.225^oC=i\times 1.86^oC/m\times 0.100m\\\\i=\frac{0.225}{1.86\times 0.100}=1.21$

Hence, the experimental van't Hoff factor is 1.21

7 0

3 years ago