Answer:
1.4952 grams of sodium bicarbonate she would need to ingest to neutralize this much HCl.
Explanation:
Moles of hydrochloric acid = n
Volume of hydrochloric acid solution = 200.0 mL = 0.200 L
Molarity of the hydrochloric acid = 0.089 M
of HCL
According to reaction, 1 mole of HCl is neutralized by 1 mole of sodium bicarbonate.
Then 0.0178 moles of HCl wil be neutralized by :
of sodium bicarbonate
Mass of 0.0178 moles of sodium bicarbonate:
0.0178 mol × 72 g/mol = 1.4952 g
1.4952 grams of sodium bicarbonate she would need to ingest to neutralize this much HCl.
Answer:
4.8 %
Explanation:
We are asked the concentration in % by mass, given the molarity of the solution and its density.
0.8 molar solution means that we have 0.80 moles of acetic acid in 1 liter of solution. If we convert the moles of acetic acid to grams, and the 1 liter solution to grams, since we are given the density of solution, we will have the values necessary to calculate the % by mass:
MW acetic acid = 60.0 g/mol
mass acetic acid (the solute) = 0.80 mol x 60 g / mol = 48.00 g
mass of solution = 1000 cm³ x 1.010 g/ cm³ (1l= 1000 cm³)
= 1010 g
% (by mass) = 48.00 g/ 1010 g x 100 = 4.8 %
As the number of electrons added to the same principal energy level increases, atomic size generally
C. Decreases.
