A 0.080L solution of Ca(OH)2 is neutralized by 0.0293L of a 3.58 M H2CrO4 solution. What is the concentration of the Ca(OH)2 solution?

1.31 M

Explanation:

Step 1:

Data obtained from the question. This include the following:

Volume of Base (Vb) = 0.080L

Molarity of base (Mb) =..?

Volume of acid (Va) = 0.0293L

Molarity of acid (Ma) = 3.58 M

Step 2:

The balanced equation for the reaction. This is given below:

H2CrO4 + Ca(OH)2 → CaCrO4 + 2H2O

From the balanced equation above,

The mole ratio of the acid (nA) = 1

The mole ratio of the base (nB) = 1

Step 3:

Determination of the concentration of base.

The concentration of the base can be obtained as follow:

MaVa/MbVb = nA/nB

3.58 x 0.0293 / Mb x 0.080 = 1

Cross multiply

Mb x 0.080 = 3. 58 x 0.0293

Divide both side by 0.080

Mb = (3.58 x 0.0293)/0.08

Mb = 1.31 M

Therefore, the concentration of the base, Ca(OH)2 is 1.31 M

the answer is barium and copper bromide

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