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konstantin123 [22]
2 years ago
12

You need to make a 1:4 dilution of dye in water so that the total volume in test tube one is 60 ml.

Chemistry
1 answer:
elixir [45]2 years ago
8 0

In the serial dilution for the preparation of the dye, the total dilution factor is 1/20

<h3>What is dilution of solution?</h3>

Dilution is the process of preparing solutions of lesser concentration from a more concentrated solution.

In test tube 1:

Total volume = 60 mL

Dilution factor = 1:4

1 parts stock solution is required for every four parts diluent

Total parts = 5

volume of diluent = 4/5 × 60 mL = 48 mL

48 mL of diluent is added to 12 mL of stock solution to prepare the solution in test tube 1.

In test tube 2:

Total volume = 15 mL

Dilution factor = 1:5

1 parts stock solution is required for every five parts diluent

Total parts = 6

volume of diluent = 5/6 × 15 mL = 12.5 mL

12.5 mL of diluent is added to 2.5 mL of stock solution from tube 1 to prepare the solution in test tube 2.

Total dilution factor = 1/4 × 1/5 = 1/20

Learn more about dilution at: brainly.com/question/24881505

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Given the standard enthalpy changes for the following two reactions: (1) 2Fe(s) + O2(g)2FeO(s)...... ΔH° = -544.0 kJ (2) 2Zn(s)
Keith_Richards [23]

Answer:

-76.3 kJ

Explanation:

Here is the complete question

Given the standard enthalpy changes for the following two reactions:

(1) 2Fe(s) + O₂(g) → 2FeO(s)......ΔH° = -544.0 kJ

(2) 2Zn(s) + O₂(g) → 2ZnO(s)......ΔH° = -696.6 kJ. What is the standard enthalpy change for the reaction:

(3) FeO(s) + Zn(s) → Fe(s) + ZnO(s)......ΔH° = ?

Solution

Since (1) 2Fe(s) + O₂(g) → 2FeO(s)......ΔH° = -544.0 kJ

reversing the reaction, we have

2FeO(s) → 2Fe(s) + O₂(g) ......ΔH° = +544.0 kJ (4)

Adding reactions (2) and (3), we have

2FeO(s) → 2Fe(s) + O₂(g) ......ΔH° = +544.0 kJ (4)

2Zn(s) + O₂(g) → 2ZnO(s)......ΔH° = -696.6 kJ  (2)

This gives

2FeO(s) + 2Zn(s) → 2Fe(s) + 2ZnO(s)......ΔH° =

The enthalpy change for this reaction is the sum of enthalpy changes for reaction (2) and (3) = ΔH° = +544.0 kJ + (-696.6 kJ)

= +544.0 kJ - 696.6 kJ)

= -152.6 kJ

Since the required reaction is (3) which is FeO(s) + Zn(s) → Fe(s) + ZnO(s)

we divide the enthalpy change for reaction (4) by 2 to obtain the enthalpy change for reaction (3).

So, ΔH° = -152.6 kJ/2 = -76.3 kJ

So, the standard enthalpy change for the reaction

FeO(s) + Zn(s) → Fe(s) + ZnO(s) is -76.3 kJ

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3 years ago
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Vertical columns are called groups, or families. 

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Answer:

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