Answer:
HBr(aq) + LiOH(aq) = LiBr(aq) + H2O(l)
Explanation:
For this reaction, the reactants are the hydrobomic acid and the lithium hydroxide which produces the products lithium bromide and water.
Answer:
AgNO3 -soluble in water
AgCl- insoluble in water
AgI- Insoluble in water
Explanation:
The solubility of chemical compounds in water is easily predicted by a set of rules generally referred to as the solubility rules. These rules are usually based on experimental observation of diverse groups of compounds.
According to the solubility rules, nitrates are soluble in water including the nitrates of silver. However, halides are soluble in water except those of silver, lead and mercury.
<span>LiOH+HBr---> LiBr +h20. Moles of LiOH = 10/24 = 0.41moles. According to stoichiometry, moles of LiOH = moles of LiBr = 0.41moles. Therefore mass of LiBr =moles of LiBr x molecular weight of LiBr = o.41 x 87 = 35.67g. Hope it helps </span>
Answer:
B) 0.32 %
Explanation:
Given that:

Concentration = 1.8 M
Considering the ICE table for the dissociation of acid as:-

The expression for dissociation constant of acid is:
![K_{a}=\frac {\left [ H^{+} \right ]\left [ {CH_3COO}^- \right ]}{[CH_3COOH]}](https://tex.z-dn.net/?f=K_%7Ba%7D%3D%5Cfrac%20%7B%5Cleft%20%5B%20H%5E%7B%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20%7BCH_3COO%7D%5E-%20%5Cright%20%5D%7D%7B%5BCH_3COOH%5D%7D)


Solving for x, we get:
<u>x = 0.00568 M</u>
Percentage ionization = 
<u>Option B is correct.</u>