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.
V = nRT/P
V = 0.685 mol*(.0821 L*atm/K*mol)*273 K/1 atm
Answer:
Speed is measured over time.
Explanation:
Answer:
1. CaCO3 + 2HCl → CaCl2 + H2O + CO2
2. C6H12O2 + 8O2 → 6CO2 + 6H2O
Explanation: