Answer is: <span>he boiling point of a 1.5 m aqueous solution of fructose is </span>100.7725°C.
The boiling point
elevation is directly proportional to the molality of the solution
according to the equation: ΔTb = Kb · b.<span>
ΔTb - the boiling point
elevation.
Kb - the ebullioscopic
constant. of water.
b - molality of the solution.
Kb = 0.515</span>°C/m.
b = 1.5 m.
ΔTb = 0.515°C/m · 1.5 m.
ΔTb = 0.7725°C.
Tb(solution) = Tb(water) + ΔTb.
Tb(solution) = 100°C + 0.7725°C = 100.7725°C.
Answer:
elements in same group have same valence electron.
period of element is equal to valence shell
Explanation:
if the elements are in a same group then they will be having same number of valence electron.
the period of an element in periodic table is equal to the valence shell of the element. that is if the valence electron are in 3 rd shell then the element will be in third period.
<em>Answer:</em>
- 0.052301 km have 5 significant figure
- 400 cm have 1 significant figure
- 50.0 m have 3 significant figure
- 4500.01 ml have 6 significant figure
<em>Explanation:</em>
According to rules of significant figure
0.052301 km have 5 significant figure:
- Zero to the left of the first non zero digit not significant.
- Zero between the non zero digits are significant.
<em>400 cm have 1 significant figure:</em>
- Trailing zeros are not significant in numbers without decimal points.
<em>50.0 m have 3 significant figure:</em>
- Trailing zeros are significant in numbers when there is decimal points.
<em>4500.01 ml have 6 significant figure:</em>
- Zero between the non zero digits are significant.
Answer:
The density of ozone is 4.24.
Explanation:
The relation between the relative rate of diffusion and density is given by :
The given ratio of the relative rate of diffusion of ozone as compared to chlorine is 6:3.
Let the density of ozone is d₂.
So, the density of ozone is 4.24.
If your choices are the following:
A. 1-inch IMC.
B. 1-inch rigid conduit.
C. 3/4 inch IMC.
D. 1/2 inch EMT.
Then the answer is C.
