Answer:
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Explanation:
may i please have a branlliests
Answer:
c) 3,3-dimethylpent-1-yne
Answer:
The answer to your question is 25.2 g of acetic acid.
Explanation:
Data
[Acetic acid] = 0.839 M
Volume = 0.5 L
Molecular weight = 60.05 g/mol
Process
1.- Calculate the number of moles of acetic acid
Molarity = moles / volume
-Solve for moles
moles = Molarity x volume
-Substitution
moles = (0.839)(0.5)
-Result
moles = 0.4195
2.- Calculate the mass of acetic acid using proportions and cross multiplications
60.05 g ----------------------- 1 mol
x ----------------------- 0.4195 moles
x = (0.4195 x 60.05) / 1
x = 25.19 g
3.- Conclusion
25.2 g are needed to prepare 0.500 L of Acetic acid 0.839M
B) Mg is the alkaline earth metal w/12 protons so following the periodic table to the halogen in the same period is
Cl: Chlorine
C) The Neutral noble has w/ 18 electrons is argon so the metal in the same row is
Na: Sodium
6 because after the 'H' is the number '6'
