# 2mL of a serum sample was added to 18mL of phosphate buffered saline (PBS) in Tube 1. 10mL of Tube 1 was added to 40mL of PBS in Tube What is the dilution of serum in Tube 2?

Tube 2 has a total dilution of 1:50

Explanation:

We have a 2 ml serum sample added to a 18 mL phosphate buffered saline sample in tube 1. This means now in tube 1 there is 20 mL.

We have a 1:10 (= 2:20) dilution here.

10 ml of this 1:10 diluted tube 1 is taken and added to a 40 mL of PBS in tube 2.

Now we have 50 mL in tube 2.

This is a 10:50 (= 1:5) dilution.

The total dilution is 10x5 = 50

So the total ditultion has a rate 1:50

Tube 2 has a total dilution of 1:50

## Related Questions

Identify the spectator ions in the following molecular equation. KBr(aq) + AgNO3(aq) → AgBr(s) + KNO3(aq)
I'm not sure what the rules are for assigning oxidation numbers.

The identification of the spectator ions should be K+ and NO3-

Identification of the  spectator ions:

Here the chemical equation should be written again in the net ionic equation

KBr(aq) + AgNO3(aq) → AgBr(s) + KNO3(aq)

K+ + Br- + Ag + + NO3-  = AgBr + K + NO3-

Here only aqueous solutions should be dissociated. Also, Spectator ions should be present in both sides, so these are K+ and NO3-. The rules of assigning oxidation numbers are to identify the number of valence

We can rewrite the equation KBr(aq) + AgNO3(aq) → AgBr(s) + KNO3(aq) into its net ionic equation into
K+ + Br- + Ag + + NO3-  = AgBr + K + NO3-only aqueous solutions can dissociate. Spectator ions are present in both sides, hence these are K+ and NO3-. THe rules of assigning oxidation numbers is to identify the number of valence electrons of elements and may be arbitrary depending on the charge of the molecule.

What might cause a prolonged period of heavy thunderstorm rains over relatively flat terrain

Are you asking what might a period of heavy rain over a flat terrain might cause? If yes, flash flooding.

What are the three particles of an atom? A. Neutron, electron, and ion B. Proton, neutron, and electron C. Electron, proton, and isotope D. Proton, ion, and valencium

Its b. proton, neutron, and electron

) If 23 g of FeCl2 reacts with 41 grams of Na3PO4, what is the limiting reagent? How much NaCl can be formed?

Answer: The limiting reagent is iron (II) chloride and the mass of sodium chloride formed is 21.2 grams

Explanation:

To calculate the number of moles, we use the equation:

.....(1)

• For iron (II) chloride:

Given mass of iron (II) chloride = 23 g

Molar mass of iron (II) chloride = 126.8 g/mol

Putting values in equation 1, we get:

• For sodium phosphate:

Given mass of sodium phosphate = 41 g

Molar mass of sodium phosphate = 164 g/mol

Putting values in equation 1, we get:

The chemical equation for the reaction of iron (II) chloride and sodium phosphate follows:

By Stoichiometry of the reaction:

3 moles of iron (II) chloride reacts with 2 moles of sodium phosphate

So, 0.181 moles of iron (II) chloride will react with = of sodium phosphate

As, given amount of sodium phosphate  is more than the required amount. So, it is considered as an excess reagent.

Thus, iron (II) chloride is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 moles of iron (II) chloride produces 6 mole of sodium chloride.

So, 0.181 moles of iron (II) chloride will produce = of sodium chloride.

Now, calculating the mass of sodium chloride from equation 1, we get:

Molar mass of sodium chloride = 58.5 g/mol

Moles of sodium chloride = 0.362 moles

Putting values in equation 1, we get:

Hence, the limiting reagent is iron (II) chloride and the mass of sodium chloride formed is 21.2 grams