Answer:
74.0 g/mol
Explanation:
Step 1: Write the generic neutralization reaction
HA + NaOH ⇒ NaA + H₂O
Step 2: Calculate the reacting moles of NaOH
At the equivalence point, 33.83 mL of 0.115 M NaOH react.
0.03383 L × 0.115 mol/L = 3.89 × 10⁻³ mol
Step 3: Calculate the moles of HA that completely react with 3.89 × 10⁻³ moles of NaOH
The molar ratio of HA to NaOH is 1:1. The reacting moles of HA is 1/1 × 3.89 × 10⁻³ mol = 3.89 × 10⁻³ mol.
Step 4: Calculate the molar mass of the acid
3.89 × 10⁻³ moles of HA have a mass of 0.288 g.
M = 0.288 g / 3.89 × 10⁻³ mol = 74.0 g/mol
False... Honey is greater in velosity then water
Answer:
In which [Ag+] in negligibly small and the concentration of each reactant is 1.0 M
The answer is A) PO43- < NO3- < Na+
Explanation:
Ag+ is removed from the solution just like PO43-, so there are just 2 possible answers at this point: a or b. Then we can notice that Na3PO4 releases 3 moles of Na+ and just 1 mole of NO3-
We have 100mL of each reactant with the same concentration for both (1.0 M) so:
(0.1)(1)(3)= 0.3 mol Na+
(0.1)(1)= 0.1 mol NO3-
so PO43- < NO3- < Na+
There are three variables independent, dependent ,and controlled
<u>Given information:</u>
Mass of NaCl (m) = 87.75 g
Volume of solution (V) = 500 ml = 0.5 L
Molar mass of NaCl (M) = 58.44 g/mol
<u>To determine:</u>
The molarity of NaCl solution
<u>Explanation:</u>
Molarity is defined as the number of moles of solute(n) dissolved per liter of solution (V)
i.e. M = moles of solute/liters of solution = n/V
Moles of solute (n) = mass of solute (m)/molar mass (M)
moles of NaCl = 87.75 g/58.55 g.mol-1 = 1.499 moles
Therefore,
Molarity of NaCl = 1.499 moles/0.5 L = 2.998 moles/lit ≅ 3 M
<u>Ans: (D)</u>
