Calculate the volume of 0.0500 M H2SO4 solution required to completely react with 20.0 mL of 1.00 x 10-3 M NaOH.

vol 0.05M H₂SO₄ = 0.200 ml

Explanation:

H₂SO₄ + 2NaOH => NaCl + 2 H₂O

moles = Molarity X Volume

From balanced equation 1 mole H₂SO₄ neutralizes 2 moles NaOH.

Using neutralization equation

Moles NaOH neutralized = 2 x Moles H₂SO₄ used

∴ 1(Molarity x Volume Base) = 2(Molarity x Volume Acid)

=> Volume of Acid needed = Molarity x Volume of Base / 2 x Molarity of Acid

=> Volume of Acid needed = (0.001M x 20 ml) / (2 x 0.05M) = 0.20 ml Acid

we need 0.2 mL of a 0.0500M  H2SO4 solution

Explanation:

Step 1: Data given

Molarity of H2SO4 = 0.0500 M

Volume of NaOH = 20.0 mL = 0.020 L

Molarity of NaOH = 1.00 * 10^-3 M

Step 2: The balanced equation

2NaOH + H2SO4 → Na2SO4 + 2H2O

Step 3: Calculate the volume of H2SO4

b*Ca*Va = a*Cb*Vb

⇒with b = the coefficient of H2SO4 = 1

⇒with Ca = the concentration of NaOH = 1.00 * 10^-3 M

⇒with Va = the volume of NaOH = 0.020 L

⇒with a = the coefficent of NaOH = 2

⇒with Cb = the concentration of H2SO4 = 0.0500 M

⇒with Vb = the volume of H2SO4 = TO BE DETERMINED

1*10^-3 * 0.020 = 2*0.0500 * Vb

Vb = 0.0002 L = 0.2 mL

we need 0.2 mL of a 0.0500M  H2SO4 solution

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