A) The limiting reactant is Al
b) Br2 is the excess reactant
c) The amount moles of AlBr3 that get produced will be equal to the number of moles of Al to begin with.
d) 0
Answer:
Percentage dissociated = 0.41%
Explanation:
The chemical equation for the reaction is:
The ICE table is then shown as:
Initial (M) 1.8 0 0
Change (M) - x + x + x
Equilibrium (M) (1.8 -x) x x
![K_a = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}](https://tex.z-dn.net/?f=K_a%20%20%3D%20%5Cfrac%7B%5BC_3H_6ClCO%5E-_2%5D%5BH%5E%2B%5D%7D%7B%5BC_3H_6ClCO_2H%5D%7D)
where ;
Since the value for 
is infinitesimally small; then 1.8 - x ≅ 1.8
Then;
Dissociated form of 4-chlorobutanoic acid = 
Percentage dissociated = 
Percentage dissociated = 
Percentage dissociated = 0.4096
Percentage dissociated = 0.41% (to two significant digits)
Molar mass Li2CO3 = 73.89 g/mol
Molar mass Li = 6.94g/mol Li = 6.94*2 = 13.88g
% LI = 13.88/73.89*100 = 18.78% perfectly correct.
Answer:
17.5609g
Explanation:
According to the question, a sample of mass 6.814 grams is added to another sample weighing 0.08753 grams. That is weight of sample 1 + weight of sample 2;
6.814 + 0.08753 = 6.90153grams
Next, the subsequent mixture is then divided into exactly 3 equal parts i.e. 6.90153grams divided by 3
= 6.90153/3
= 2.30051grams.
One of the equal parts is 2.30051grams, which is then multiplied by 7.6335 times I.e. 2.30051 × 7.6335 = 17.5609grams
Therefore, the final mass is 17.5609grams
