C = n/V
n = C×V
n = 4,41M × 1,25L
n = 5,5125 mol
mKI: 39+127 = 166 g/mol
1 mol --------- 166g
5,5125 mol --- X
X = 166×5,5125 = 915,075g KI
:)
Answer: 4.4
Explanation:
Thats what my computer says after i submit answer
Answer:
2.76 × 10⁻¹¹
Explanation:
I don’t have access to the ALEKS Data resource, so I used a different source. The number may be different from yours.
1. Calculate the free energy of formation of CCl₄
C(s)+ 2Cl₂(g)→ CCl₄(g)
ΔG°/ mol·L⁻¹: 0 0 -65.3
ΔᵣG° = ΔG°f(products) - ΔG°f(reactants) = -65.3 kJ·mol⁻¹
2. Calculate K

T = (25.0 + 273.15) K = 298.15 K

NO, it should not be the process is still the same . the only factors about it that should change is the experiment itself.:)
Answer:
Kb = 

Explanation:
For a weak organic base, the formula to find
is given by:

where c is the concentration of base.
Here c= 

Substituting the above values in the formula,we get:

Hence:
Kb = 
