You are given a sample of blood for analysis and the experimental protocol requires the following steps: 1. pipette 0.5 ml of blood into 1.0 ml of saline in a fresh test tube. 2. remove 0.05 ml of this sample and add to 1.0 ml of the assay reagent in a fresh test tube what is to total dilution of the blood after step 2 in the protocol?

The blood is diluted to 4.9 folds.

Explanation:

Steps: 1. Pipette 0.5 mL of blood into 1.0 mL of saline in a fresh test tube.

Initial volume of the Volume of blood before any dilution ,V= 0.5 mL

Volume of saline = 1.0 mL

Volume of the solution after mixing = 0.5 mL + 1.0 mL = 1.5 mL

Step : 2. Remove 0.05 mL of this sample and add to 1.0 mL of the assay reagent in a fresh test tube.

On removal of 0.05 mL from 1.5 mL the volume becomes:

= 1.5 mL - 0.05 mL = 1.45 mL

And 1.0 ml of the essay reagent is added in to 1.45 mL of the solution obtained just above:

Volume of the blood solution = 1.45 mL + 1.0 mL = 2.45 mL

Final volume of the blood solution after dilution = ,V'= 2.45 mL

Fold - dilution =

V =Volume of original solution before dilution

V' = Total Volume of solution after dilution

The blood is diluted to 4.9 folds.

"record the time to complete a chemical reaction" would most likely be part of a lab procedure.

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