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 ————
60.8 g ammonia
Answer:
4180J
Explanation:
(25.0g)(4.184J/g°C)(75°C-35.0°C)
(25.0g)(40.0°C)(4.184J/g°C)
(1.00*10³g°C)(4.184J/g°C) = 4184J
use sig figs:
4180J