A) Combustion
Hope this helps
Answer:
a) 8.33 ml of the original stomach acid is neutralized
b) 191.67 ml of the stomach acid was neutralized
c) 249.68 ml acid would be neutralized by the original tablet
Explanation:
a) how much of the stomach acid had been neutralized in the 25 mL sample wich was titrated?
25.5 ml of a NaOH solution is equivalent to 25.00 ml of the original stomach acid
8.5 ml NaOH * (25.00 ml original stomach acid / 25.5 ml NaOH) = 8.33 ml original stomach acid
b) how much stomach acid was neutralized y the 4.3628 g tablet?
It takes 8.5 ml NaOH to neutralize 8.33 ml original acid (this is the answer for question 1)
This means the antacid neutralized = 200 ml - 8.33 ml = 191.67 ml
c) how much stomach acid would have been neutralized by the original 5.6832 g tablet
4.3628 g antacid is equivalent to 191.67 ml acid ( this is the answer for question 2)
5.6832g antacid * (191.67 ml acid / 4.3628 g antacid) = 249.68 ml acid
<span>B) N2(g) + 3H2(g) → 2NH3(g)
</span>I hope this helps :)
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Answer:
The sum of the molar masses of each isotope of the element.
