The correct classification of the solutes are as follows:
<span>hydroiodic acid hi = strong electrolyte
calcium hydroxide ca(oh)2 = weak electrolyte
hydrofluoric acid hf = weak electrolyte
methyl amine ch3nh2 = weak electrolyte
sodium bromide nabr = strong electrolyte
propanol c3h7oh = non electrolyte
sucrose c12h22o11 = non electrolyte
Strong electrolytes are substances that completely ionizes in aqueous solution while weak electrolytes are those that partially ionizes. Non electrolytes are substances that cannot conduct electric charge since there are no ions in the solution.</span>
Answer:
from the pic:
since molar mass is 118.084 g/mol;
It is an aldehyde with structure ( second pic )
<span> 52.0ml of 0.35M CH3COOH : 0.052 L(0.35M) = .0182 mol of CH3COOH.
</span>
<span>31.0ml of 0.40M NaOH : .031 L(0.40M) = .0124 mol of NaOH.
</span>
<span>After the reaction, .0124 Mol CH3COO- is generated and .058 mol CH3COOH is left un-reacted. The concentration would be 12.4/V and 5.8/V, respectively. Therefore:
</span>
<span>pH = -log([H+]) = -log(Ka*[CH3COOH]/[CH3COO-]) </span>
<span>= -log(1.8x10^-5*5.8/12.4) = 5.07</span>
We can use the dilution formula to find the volume of the diluted solution to be prepared
c1v1 = c2v2
Where c1 is concentration and v1 is volume of the concentrated solution
And c2 is concentration and v2 is volume of the diluted solution to be prepared
Substituting the values in the equation
15 M x 25 mL = 3 M x v2
v2 = 125 mL
The 25 mL concentrated solution should be diluted with distilled water upto 125 mL to make a 3 M solution
Answer: The bold staircase in the periodic table allows us to classify which elements are metalloids.
Explanation: Additionally, it acts like a "divider" that allows us to properly distinguish the metals from the non-metals in the periodic table.
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