Achemist needs to create a series of standard cu2+(aq)cu2+(aq) solutions for an absorbance experiment. for the first standard, he uses a pipet to transfer 5.005.00 ml of a 2.172.17 m cu2+(aq)cu2+(aq) stock solution to a 500.0500.0 ml volumetric flask, and he adds enough water to dilute to the mark. he then uses a second pipet to transfer 25.0025.00 ml of the second solution to a 100.0100.0 ml volumetric flask, and he adds enough water to dilute to the mark. calculate the concentration of the cu2+(aq)cu2+(aq) solution in the 100.0100.0 ml volumetric flask.

The concentration of the Cu²⁺ solution in the 10,0 mL volumetric flask is 5,425×10⁻³ M Cu²⁺

Explanation:

The first dilution from a 2,17M Cu²⁺ solution gives:

2,17 M Cu²⁺ × = 0,0217 M  Cu²⁺solution

The second dilution gives as concentration:

0,0217 M Cu²⁺ × = 5,425×10⁻³ M Cu²⁺ solution

I hope it helps!

5.43 × 10⁻³ M

Explanation:

A chemist needs to create a series of standard Cu²⁺(aq) solutions for an absorbance experiment. For the first standard, he uses a pipet to transfer 5.00 mL of a 2.17 M Cu²⁺(aq) stock solution to a 500.0 mL volumetric flask, and he adds enough water to dilute to the mark. He then uses a second pipet to transfer 25.00 mL of the second solution to a 100.0 mL volumetric flask, and he adds enough water to dilute to the mark. Calculate the concentration of the Cu²⁺(aq) solution in the 100.0 mL volumetric flask.

Two successive dilutions are performed. In each one, we will use the dilution rule.

M₁ × V₁ = M₂ × V₂

where,

M: molarity

V: volume

1: initial solution (concentrated)

2: final solution (diluted)

First dilution

M₁ × V₁ = M₂ × V₂

2.17 M × 5.00 mL = M₂ × 500.0 mL

M₂ = 0.0217 M

This is the initial concentration for the second dilution.

Second dilution

M₁ × V₁ = M₂ × V₂

0.0217 M × 25.00 mL = M₂ × 100.0 mL

M₂ = 5.43 × 10⁻³ M

2b2t

explanation:

2b2t

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2