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.
Ok thanks for the valuble info.
The grams of NaCl that are required to make 150.0 ml of a 5.000 M solution is 43.875 g
calculation
Step 1:calculate the number of moles
moles = molarity x volume in L
volume = 150 ml / 1000 = 0.15 L
= 0.15 L x 5.000 M = 0.75 moles
Step 2: calculate mass
mass = moles x molar mass
molar mass of NaCl = 23 + 35.5 = 58.5 mol /L
mass is therefore =0.75 moles x 58.5 mol /l =43.875 g
Notice q=3/2, is half of the original q = 3(<span>1/2</span>)<span>t/28.8
your welcome
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