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
869.6 × 10¹⁴ molecules of EDB
Explanation:
We have 1.9 lb of flour with a EDB concentration of 31.5 ppb.
We need to transform lb in grams.
1 lb = 453.6 grams
1.9 lb = (1.9 × 453.6) / 1 = 861.8 grams
Now we determine the number of molecules of EDB in the sample by devise the following reasoning:
if we have 31.5 × 10⁻⁹ g of EDB in 1 g of sample
then we have X g of EDB in 861.8 g of sample
X = (31.5 × 10⁻⁹ × 861.8) / 1 = 27146.7 × 10⁻⁹ g of EDB
Molecular mass of EDB (C₂H₄Br₂) = 188 g/mole
Taking in account that 1 mole of any substance contains 6.022 × 10²³ (Avogadro’s number) molecules we devise the following reasoning:
if 188 g of EDB contains 6.022 × 10²³ molecules
then 27146.7 × 10⁻⁹ g of EDB contains Y molecules
Y = (27146.7 × 10⁻⁹ × 6.022 × 10²³) / 188 = 869.6 × 10¹⁴ molecules of EDB
Learn more:
about Avogadro’s number
brainly.com/question/1445383
#learnwithBrainly
Answer:
The answer to your question is 0.22
Explanation:
Data
Acetonitrile (CH₃CN) density = 0.786 g/ml
Methanol (CH₃OH) density = 0.791 g/ml
Volume of CH₃OH = 22 ml
Volume of CH₃CN = 98.4 ml
Process
1.- Calculate the mass of Acetonitrile and the mass of Methanol
density = mass/ volume
mass = density x volume
Acetonitrile
mass = 0.786 x 98.4
= 77.34 g
Methanol
mass = 0.791 x 22
= 17.40 g
2.- Calculate the moles of the reactants
Acetonitrile molar mass = (12 x 2) + (14 x 1) + (3 x 1)
= 24 + 14 + 3
= 41 g
Methanol molar mass = (12 x 1) + (4 x 1) + (16 x 1)
= 12 + 4 + 16
= 32 g
Moles of Acetonitrile
41 g ----------------- 1 mol
77.34g ------------ x
x = (77.34 x 1) / 41
x = 1.89 moles
Moles of Methanol
32 g -------------- 1 mol
17.40 g --------- x
x = (17.40 x 1)/32
x = 0.54 moles
3.- Calculate the mole fraction of Methanol
Total number of moles = 1.89 + 0.54
= 2.43
Mole fraction = moles of Methanol / total number of moles
Mole fraction = 0.54/ 2.43
Mole fraction = 0.22
