Answer:
Van't Hoff factor for AlCl₃ = 3 (Approx)
Explanation:
Given:
Number of observed particular = 1.79 M
Number of theoretical particular = 0.56 M
Find:
Van't Hoff factor for AlCl₃
Computation:
Van't Hoff factor for AlCl₃ = Number of observed particular / Number of theoretical particular
Van't Hoff factor for AlCl₃ = 1.79 M / 0.56 M
Van't Hoff factor for AlCl₃ = 3.19
Van't Hoff factor for AlCl₃ = 3 (Approx)
It releases energy (what I am saying in other words that "D. Releases" would be your answer)
Answer:
2K + Br2 --> 2KBr is the answer
Answer:
ΔHrxn = - 1534.3 J
Explanation:
Given the assumptions and the formula for the change in enthalpy:
ΔHrxn = m x C x ΔT, where
m is the mass of solution given 135.4 g
C is the heat capacity 4.2 J/g .K and,
ΔT is the change in temperature
we have ,
T₁ = ( 18.1 + 273) K = 291.1 K
T₂ = ( 15.4 +273) K = 288.4 K
ΔHrxn = 135.3 g x 4.2 J/gK x ( 288.4 -291.1 ) K = - 1534.3 J
After verifying our result has the correct unit, the answer is -1534.3 Joules, and the negative sign tells us it is an endothermic reaction decreasing the final temperature.
194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.
Explanation:
In order to convert the given number of molecules of BCl₃ to grams, first we have to convert the molecules to moles.
It is known that 1 moles of any element has 6.022×10²³ molecules.
Then 1 molecule will have 
moles.
So 
Thus, 1.66 moles are included in BCl₃.
Then in order to convert it from moles to grams, we have to multiply it with the molecular mass of the compound.
As it is known as 1 mole contains molecular mass of the compound.
As the molecular mass of BCl₃ will be
Mass of boron is 10.811 g and the mass of chlorine is 35.453 g.
Molar mass of BCl₃ = 10.811+(3×35.453)=117.17 g.
So, 194.5 g of BCl₃ is present in 1 × 10²⁴ molecules of BCl₃.
