A chemistry student weighs out 0.0617g of acetic acid (HCH3CO2) into a 250.mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1300 M NaOH solution. Calculate the volume of solution the student will need to add to reach the equivalence point. Round your answer to 3 significant digits.

We have to add 7.92 mL of NaOH

Explanation:

Step 1: Data given

Mass of acetic acid = 0.0617 grams

Volume of the flask = 250.0 mL = 0.250 L

Molarity of NaOH = 0.1300 M

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles acetic acid

Moles acetic acid = mass acetic acid / molar mass acetic acid

Moles acetic acid = 0.0617 grams / 60.05 g/mol

Moles acetic acid = 0.00103 moles

Step 4: Calculate concentration CH3COOH

Concentration CH3COOH = moles CH3COOH / volume

Concentration CH3COOH = 0.00103 moles / 0.250 L

Concentration CH3COOH = 0.00412 M

Step 5: Calculate the volume of NaOH needed

C1*V1 = C2*V2

⇒with C1 = the concentration of CH3COOH = 0.00412 M

⇒with V1 = the volume of CH3COOH = 0.250 L

⇒with C2 = the concentration of NaOH = 0.130 M

⇒with V2 = the volume of NaOH needed = TO BE DETERMINED

0.00412 M * 0.250 L = 0.130 M * V2

V2 = (0.00412 M * 0.250 L ) / 0.130 M

V2 = 0.00792 L = 7.92 mL

We have to add 7.92 mL of NaOH

it would be d. henry moseley

hope this !

and sorry if i'm late. have a good rest of your day! : d

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally