Answer:
a, and f.
Explanation:
To be deprotonated, the conjugate acid of the base must be weaker than the acid that will react, because the reactions favor the formation of the weakest acid. The pKa value measures the strength of the acid. As higher is the pKa value, as weak is the acid. So, let's identify the conjugate acid and their pKas:
a. NaNH2 will dissociate, and NH2 will gain the proton and forms NH3 as conjugate acid. pKa = 38.0, so it happens.
b. NaOH will dissociate, and OH will gain the proton and forms H2O as conjugate acid. pKa = 14.0, so it doesn't happen.
c. NaC≡N will dissociate, and CN will gain a proton and forms HCN as conjugate acid. pKa = 9.40, so it doesn't happen.
d. NaCH2(CO)N(CH3)2 will dissociate and forms CH3(CO)N(CH3)2 as conjugate acid. pKa = -0.19, so it doesn't happen.
e. H2O must gain one proton and forms H3O+. pKa = -1.7, so it doesn't happen.
f. CH3CH2Li will dissociate, and the acid will be CH3CH3. pKa = 50, so it happens.
In titration, the moles of acid equal moles of base. You were given that 22.75ml of 0.215M NaOH is used, so calculate the number of moles of that base the experiment used in total. After that because you know mol base = mol acid, whatever amount of base you use must be the total amount of acid present in the solution. You were given the volume of the acid, and you have just found the total mols of acid. Using these two information, solve for the concentration. And one more thing, even though I'm pretty sure it won't affect your answer, you should always convert things to the proper units. Since the concentration we're talking about in this problem is molarity, which has the unit mol/L, you should always have all of your numbers in these units. It just make it simpler and will not confuse you
Exposure to small amounts of lead<span> over a long period of time is called chronic toxicity. </span>Lead<span> is particularly </span>dangerous<span> because once it gets into a person's system, it is distributed throughout the body just like helpful minerals such as iron, calcium, and zinc. And </span>lead<span> can cause harm wherever it lands in the body.</span>
Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
M(Ca(NO3)2)= M(Ca) + M(N) + 6M(O)= 40.0 +14.0 +6*16.0 = 150 g/mol
15.0 g Ca(NO3)2 * 1mol/150 g = 0. 100 mol Ca(NO3)2
Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
1 mol 2 mol
0.100 mol 0.200 mol
We have 0.2 mol NO3⁻ in 300. mL=0.300 L of solution,
so
0.200 mol NO3⁻ / 0.300 L solution ≈ 0.667 mol NO3⁻ /L solution = 0.667 M
Concentration of NO3⁻ is 0.667 M.
I think the answer would be trenches but I’m sorry if I’m wrong
