The mole ratio of acetic acid to water in 100 g of vinegar is 0.015 : 0.985.
<h3>What is the mole ratio of acetic acid to water in 100 g of vinegar?</h3>
The mole ratio of acetic acid to water in 100 g of vinegar is determined from their percentage composition.
The percentage composition of acetic acid and water in vinegar is 5% acetic acid and 95% water.
In 100 g of vinegar, there are 5 g of acetic acid and 5 g of water.
Moles = mass/molar mass
molar mass of acetic acid = 62 g/mol
molar mass of water = 18 g/mol
moles of vinegar = 5/62 = 0.08
moles of water = 95/18 = 5.28
total moles = 5.36
Mole ratio of vinegar to water = 0.08/5.36 : 5.28/5.36
Mole ratio of vinegar to water = 0.015 : 0.985
In conclusion, the mole ratio is determined from the percentage composition of acetic acid and water in vinegar.
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Answer:
We are given:
Volume (V) = 0.25 L
Pressure (P) = 0.93 atm
Temperature (T) = 15.4°C OR 288.4 K
<u>Solving for the number of moles of CO₂:</u>
From the ideal gas equation:
PV = nRT
replacing the variables
0.93 * 0.25 = n (0.082)(288.4)
n = 0.00983 moles
<u>Number of molecules:</u>
Number of moles= 0.00983
number of molecules in 1 mole = 6.022 * 10²³
Number of molecules in 0.00983 moles = 0.00983 * 6.022 * 10²³
Number of molecules = 5.91 * 10²¹
Answer:
molarity of diluted solution = 1.25 M
Explanation:
Using,
C1V1 (Stock solution) = C2V2 (dilute solution)
given that
C1 = 2.50M
V1 = 250ML
C2 = ?
V2 = 500ML
2.50 M x 250 mL = C2 x 500 mL
C2 = (2.50 M x 250 mL) / 500 mL
C2 = 1.25 M
Hence, molarity of diluted solution = 1.25 M
Answer:
sorry i'm not a scientist but i think it's C
Explanation:
sorry again if i get it wrong for u :(
Answer:
Explanation:
In a chemical change through chemical reactions, a new kind of matter forms. When a paint is mixed together, a physical change has occurred and not a chemical change. The different color one perceives is just a mere overlap between the reflected white and red color which produces another hue.
Not all color changes infers a chemical reaction. A paint is a complex mixture on its own.
