Answer:
a. Rate = k×[A]
b. k = 0.213s⁻¹
Explanation:
a. When you are studying the kinetics of a reaction such as:
A + B → Products.
General rate law must be like:
Rate = k×[A]ᵃ[B]ᵇ
You must make experiments change initial concentrations of A and B trying to find k, a and b parameters.
If you see experiments 1 and 3, concentration of A is doubled and the Rate of the reaction is doubled to. That means a = 1
Rate = k×[A]¹[B]ᵇ
In experiment 1 and to the concentration of B change from 1.50M to 2.50M but rate maintains the same. That is only possible if b = 0. (The kinetics of the reaction is indepent to [B]
Rate = k×[A][B]⁰
<h3>Rate = k×[A]</h3>
b. Replacing with values of experiment 1 (You can do the same with experiment 3 obtaining the same) k is:
Rate = k×[A]
0.320M/s = k×[1.50M]
<h3>k = 0.213s⁻¹</h3>
Answer: The freezing point depression is
Explanation:
Depression in freezing point is given by:
= Depression in freezing point
= freezing point constant =
m= molality =
Thus freezing point depression is
Answer:
1. How many ATOMS of boron are present in 2.20 moles of boron trifluoride? atoms of boron.
2. How many MOLES of fluorine are present in of boron trifluoride? moles of fluorine.
Explanation:
The molecular formula of boron trifluoride is .
So, one mole of boron trifluoride has one mole of boron atoms.
1. The number of boron atoms in 2.20 moles of boron trifluoride is 2.20 moles.
The number of atoms in 2.20 moles of boron is:
One mole of boron has ---- atoms.
Then, 2.20 moles of boron has
-
2. Calculate the number of moles of BF3 in 5.35*1022 molecules.
One mole of boron trifluoride has three moles of fluorine atoms.
Hence, 0.0888moles of BF3 has 3x0.0888mol of fluorine atoms.
=0.266mol of fluorine atoms.
<span>362.51 Kelvin
ln (p1/p2) =( dH / R) (1/T2 - 1/T1)
ln (760 Torr /520Torr) =( 40,700 Joules / 8.314 J molâ’1K-1)(1/T2 - 1/373K)
ln (1.4615) =( 4895.35)(1/T2 - 0.002681)
0.37946 = 4895.35/T2 - (0.002681)(4895.35)
0.37946 = 4895.35/T2 - (13.124)
0.37946 + 13.124 = 4895.35/T2
13.5039 = 4895.35/T2
T2 = 4895.35 / 13.5039
T2 = 362.51
answer is 362.51 Kelvin
- 273
answer is also 89.5 Celsius</span>
21.69mL - 20.70mL = .99mL Fe
7.8 g / .99 mL = 7.9g/mL