Answer: Bromide is many orders of magnitude better than fluoride in leaving group ability
Explanation:
As Size of an atom Increases, the Basicity Decreases this is because if we move downwards from the top of the periodic table to the bottom of the periodic table, the size of an atom increases. As size increases, basicity will decrease, meaning the element will be less likely to act as a base implying that the element will be less likely to share its electrons.
in the same vein. With an increase in size, basicity decreases, making the ability of the leaving group to leave increase to increase . This can be seen in the halogens going down the group from
F--- worst
Cl----fair
Br ----good
I-----excellent
with fluorine having the worst ability to leave than Bromine which is better in terms of the leaving group ability.
Answer:
uzb-prwk-khj
indians and all other countries can join
The heat released by the substance in the calorimeter is equal to the heat absorbed by water which results to the decrease and increase in temperature, respectively.
We use m Cp ΔT to balance the heat involved
(m Cp ΔT) subs in calorimeter = <span>(m Cp ΔT) water
</span>125 g * Cp * (97.0-23.5 ) C = 250 g *(4.18 J/C g)* (23.5-20)
Cp = 0.398 J/Cg
Answer is B
Answer:
Final molarity of iodide ion C(I-) = 0.0143M
Explanation:
n = (m(FeI(2)))/(M(FeI(2))
Molar mass of FeI(3) = 55.85+(127 x 2) = 309.85g/mol
So n = 0.981/309.85 = 0.0031 mol
V(solution) = 150mL = 0.15L
C(AgNO3) = 35mM = 0.035M = 0.035m/L
n(AgNO3) = C(AgNO3) x V(solution)
= 0.035 x 0.15 = 0.00525 mol
(AgNO3) + FeI(3) = AgI(3) + FeNO3
So, n(FeI(3)) excess = 0.00525 - 0.0031 = 0.00215mol
C(I-) = C(FeI(3)) = [n(FeI(3)) excess]/ [V(solution)] = 0.00215/0.15 = 0.0143mol/L or 0.0143M
<span>B)<span>C2H6O<span>2
</span></span></span>
First, convert each percentage to grams: 38.7g, 9.70g, and 51.6g.
Next, calculate the number of moles of each element, based on the number of grams given.
C = 3.23 mol
H = 8.91 mol
O = 3.23 mol
Set up the ratio of moles of each element:
C3.34H9.70O3.23. Convert the decimals to whole numbers by dividing by the smallest subscript, 3.23.
The empirical formula is CH3O.
Now, compute the formula mass, which is 31. Finally, divide the molecular mass by the formula mass, 62/31 = 2. Multiple the subscripts by 2 to get the molecular formula.
