Answer:
freezing point (°C) of the solution = - 3.34° C
Explanation:
From the given information:
The freezing point (°C) of a solution can be prepared by using the formula:

where;
i = vant Hoff factor
the vant Hoff factor is the totality of the number of ions in the solution
Since there are 1 calcium ion and 2 nitrate ions present in Ca(NO3)2, the vant Hoff factor = 3
= 1.86 °C/m
m = molality of the solution and it can be determined by using the formula

which can now be re-written as :



molality = 0.599 m
∴
The freezing point (°C) of a solution can be prepared by using the formula:



the freezing point of water - freezing point of the solution
3.34° C = 0° C - freezing point of the solution
freezing point (°C) of the solution = 0° C - 3.34° C
freezing point (°C) of the solution = - 3.34° C
Below are some conversions.
336.715 m/s
1212.175 km/h
753.206 mph
1104.709 ft/s
654.08 knots
Try using this website:
www.sengpielaudio.com/calculator-speedsound.htm
Tue answer is 13 hope this help
Answer:
d =~ 5.8μm
d =~ 0.13 μm
Explanation:
when the doping concentrations are 5 × 10^15 cm^-3
d = v^-1/3 ; where d represent the distance between the atoms , and v represent the volume
d =1/ ∛v
d = 1/ ∛5 × 10^15
d = 1/ 170997.5
d = 5.85 × 10 ^ -6
d =~ 5.8μm
when the doping concentrations are 5 × 10^20 cm^-3
d = v^-1/3 ; where d represent the distance between the atoms , and v represent the volume
d =1/ ∛v
d = 1/ ∛5 × 10^20
using the principle of surds and standard forms, we have
d = 1/ ∛0.5 × 10^21
d = 1/7937005.26
d = 1.26 × 10 ^ -7
d = 0.126 × 10 ^ -6
d =~ 0.13 μm