If 36.2g of Acetic Acid (HC2H302) was dissolved in 300. mL of water, what is the

Chemistry

1 answer:

uysha [10]3 years ago

7 0

Answer:

2.01 M

Explanation:

Step 1: Calculate the moles of acetic acid (HC₂H₃O₂)

The molar mass of acetic acid is 60.05 g/mol. We will use this data to calculate the moles corresponding to 36.2 g of acetic acid.

$36.2g \times \frac{1mol}{60.05g} = 0.603mol$

Step 2: Convert the volume of solution to liters

We will use the relation 1000 mL = 1 L. We assume that the volume of solution is that of water (300 mL)

$300mL \times \frac{1L}{1000mL} = 0.300L$

Step 3: Calculate the molarity of the solution

The molarity is equal to the moles of solute (acetic acid) divided by the liters of solution

$M = \frac{0.603mol}{0.300L} = 2.01 M$

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How many atoms of zinc are in 0.60 mol of zinc? 2. how many moles is 3.52 × 1024 molecules of water? 3. how many atoms are in 2.

Veronika [31]

question 1
by use of Avogadro law
that is,1 mole = 6.02 x10^23 atoms
what about 0.60 moles
by use of cross multipication

=(0.60 mole/ x 6.02 x10^23)/ 1mole = 3.612 x10^23 atoms of Zn

question 2
by use of Avogadro law constant
that is 1 mole =6.02 x10^23 molecules
what about 3.52 x10^24 molecules
by cross multiplication
=( 1 mole x3.52 x10^24 molecules/6.02 x10^23 molecules) = 5.847 moles of water

question 3

by use of Avogadro law constant
1 mole = 6.02 x10^23 atoms
what about 2 moles =? atoms
by use of cross multiplication
=( 2moles x 6.02 x10^23 )/1mole= 1.204 x10^24 atoms of Li

question 4

by use of Avogadro law constant
1 mole = 6.02 x10^23 atoms
what about 6.02 x10^23 atoms =? moles
cross multiplication
(1 mole x6.02 x10^23 atoms)/(6.02 x10^23 atoms)= 1 mole of carbon

question 5
by use Avogadro law constant
1 mole =6.02 x10^23 molecules
what about 4.9 x10^23 moles =? moles
by cross multipication
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8 0

3 years ago

Calculate the ph of the solution upon the addition of 0.015 mol of naoh to the original buffer. express the ph to two decimal pl

tatiyna

You missed a lot of details in your question, so when we have the complete question as the attached picture so, the answer would be:

when we have the value of Ka = 1.8 x 10^-5 so, we can use it to get the Pka value

by using this formula:

Pka = -㏒Ka

= -㏒(1.8 x 10^-5)

= 4.7

now, after we have got the Pka we need now to get moles of NaC2H3O2 and

moles of HC2H3O2:

when moles of NaOH = 0.015 moles

when moles NaC2H3O2 after adding NaOH

= initial mol NaC2H3O2 + mol NaOH
∴moles NaC2H3O2 = 0.1 + 0.015 = 0.115 moles

and moles HC2H3O2 after adding NaOH

= initial mol HC2H3O2 - mol NaOH

∴ moles HC2H3O2 = 0.1 - 0.015 = 0.085 moles

so, when we have moles [HC2H3O2] &[NaC2H3O2] so we can substitution its values in [A] &[HA] :

by using H-H equation we can get the PH:

when PH = Pka + ㏒[A]/[HA]

PH = 4.7 + ㏒0.115/0.085

= 4.8