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2 years ago

I need help with 17,18,19,& 21

Chemistry

1 answer:

valentinak56 [21]2 years ago

3 0

Answer:

17. tie hair up

wear goggles

18. i don't know

19. you light it on the side of the box

20. it should be orange because that is the safest.

21. you turn down the electricity thing

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You need to use a dilute hydrochloric acid solution in an experiment. However, the only bottle of hydrochloric acid in your lab’

mr_godi [17]

The pH of the diluted HCl solution is 1.3.

Explanation:

Given:

The concentrated HCl solution of 8.0 M. The 1.5 mL of 8.0 M HCl is diluted with water to 250 mL volume.

To find:

The pH of the diluted HCl solution.

Solution

The concentration of the HCl solution before dilution = $M_1=8.0 M$
The volume of the HCl solution taken for dilution = $V_1=1.5 ml$
The concentration of the HCl solution after dilution = $M_2=?$
The volume of the HCl solution after dilution = $V_2=250 mL$

Using the Dilution equation:

$M_1V_1=M_2V_2\\8.0M\times 1.5ml=M_2\times 250 mL\\M_2=\frac{8.0M\times 1.5ml}{250 mL}=0.048 M$

The concentration of diluted HCl solution = 0.048 M

$HCl(aq)\rightarrow H^+(aq)+Cl^-(aq)$

In the 1 M solution of HCl, there are 1 M of hydrogen ion, then the concentration of hydrogen ions in 0.048 M of HCl will be:

$[H^+]=1\times 0.048M=0.048 M$

The pH of the diluted HCl solution :

$pH=-\log [H^+]\\=-\log [0.048M]=1.18 \approx 1.3$

The pH of the diluted HCl solution is 1.3.

Learn more about the dilution equation here:

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4 0

2 years ago

How many grams of potassium benzoate trihydrate, KC H302 - 3H20. are needed to prepare 1.5 L of a 0.120M solution of potassium b

Annette [7]

<u>Answer:</u> The mass of potassium benzoate is $8.40\times 10^{-4}g$

<u>Explanation:</u>

To calculate the mass of solute, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

We are given:

Molarity of solution = 0.120 M

Molar mass of potassium benzoate trihydrate = 214.26 g/mol

Volume of solution = 1.5 L

Putting values in above equation, we get:

$0.120M=\frac{\text{Mass of potassium benzoate trihydrate}}{214.26\times 1.5}\\\\\text{Mass of potassium benzoate trihydrate}=\frac{0.120mol/L\times 1.5L}{214.26g/mol}=8.40\times 10^{-4}g$