Answer:
Explanation:
CH₃COOH + NaOH = CH₃COONa + H₂O .
42.5 mL of .115 M of NaOH will contain .0425 x .115 moles of NaOH
= 48.875 x 10⁻⁴ moles NaOH
It will react with same number of moles of acetic acid
So number of moles of acetic acid in 3.45 mL = 48.875 x 10⁻⁴
number of moles of acetic acid in 1000 mL = 48.875 x 10⁻⁴ x 10³ / 3.45 moles
= 1.4167 moles
= 1.4167 x 60 gram
= 85 grams .
So 85 grams of acetic acid will be contained in one litre of acetic acid.
Answer:
Explanation:
Hello there!
In this case, since the reaction for the formation of ammonia is:
We can evidence the 1:2 mole ratio of nitrogen gas to ammonia; therefore, the appropriate stoichiometric setup for the calculation of the moles of the latter turns out to be:
And the result is:
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Answer : The volume of 3.0 M spinach solution added should be, 50 mL
Explanation :
Formula used :
where,
are the initial molarity and volume of spinach solution.
are the final molarity and volume of diluted spinach solution.
We are given:
Now put all the given values in above equation, we get:
Hence, the volume of 3.0 M spinach solution added should be, 50 mL
Answer:
In this section, we consider how several of the enumerated powers of Congress under the original Constitution have been interpreted. The Congress shall have Power To lay and collect Taxes, Duties, Imposts and Excises, to pay the Debts and provide for the common Defence and general Welfare of the United States.
Answer 19.9g. I’ve took the test last week at my uncle randy’s house
