if you want to remove an article from website contact us from top.

# calculate the number of moles of sulfuric acid that is contained in 0.250 ml of 0.500 m sulfuric acid solution.

Category :

### Santiago

Chicos, ¿alguien sabe la respuesta?

obtenga calculate the number of moles of sulfuric acid that is contained in 0.250 ml of 0.500 m sulfuric acid solution. de este sitio.

## Solved Calculate the number of moles of sulfuric acid that

Answer to Solved Calculate the number of moles of sulfuric acid that fuente : www.chegg.com

## Experiment 17: Concentration Units Flashcards

Study with Quizlet and memorize flashcards containing terms like molarity, molality, % (m/m) and more. ## Experiment 17: Concentration Units

Term 1 / 23 molarity

Click the card to flip 👆

Definition 1 / 23

amount of solute in moles per liter of solution, M = moles (solute) / L (solution)

Click the card to flip 👆

Created by m_sanchez0512

### Terms in this set (23)

molarity

amount of solute in moles per liter of solution, M = moles (solute) / L (solution)

molality

amount of solute in moles per kilogram of solvent, m = moles (solute) / Kg (solvent)

% (m/m)

= 100% x grams / total mass

WOTF concentration units are temperature dependent?

molarity

molarity is an example of

volumetric unit volumetric unit

value determined by volume

molality and % (m/m) are examples of

gravimetric units gravimetric unit value set by mass

In titration labs, you used weak acid hydrogen phthalate (KHC8H4)4) as a primary standard in an acid/base titration "KHP", How much of this weak acid would have to be weighed out to prepare 100.0 mL of a 0.240 M KHP solution?

...

If you actually weighed out 8.50% more KHP than required what would be the actual molarity of the solution?

...

If you used the prepared KHP solution in a lab experiment which concentration would you need to use in your calculation? Explain your reasoning.

...

You weight out 2.98 g of sodium sulfate and use it to create a 250-mL aqueous solution. What is the molarity of sodium ions in the resulting solution?

...

Calculate the number of moles of sulfuric acid that is contained in 0.250 L of 0.500 M sulfuric acid solution

Explanation - 0.250 L x 0.500 M

what mass of MgF2 is contained in 80.85 g of a 22.4% by mass solution of MgF2 in water?

Explanation - 22.4% = 100% x (x grams) / 80.85 g

sulfuric acid H2SO4 stock solution

a solution that has been prepared with known molarity

What is the molarity of a potassium bromide solution that is prepared by adding 400 mL of water to 25 mL of aqueous 2.00 M potassium bromide solution? Assume volumes are additive, report answer to three sig. figs.

...

Part A - Preparation of 100.0 mL of an aqueous solution of sucrose, C12H22O11, with specified molarity

Assigned Molarity = 0.0441 M

Volume of Solution = 100.00 mL

Mass of Sucrose = ? Answer = 1.51 g

Explanation - 0.1 L x (0.0441 M / 1 L) x (342.308 g / 1 mol)

Calculation - Mass of sucrose required

Part B - Dilution of a 1.0 M NaCl solution to a specified final volume

Assigned Final Volume = 76.2 mL

Molarity of "Stock Solution" = 1.0 M NaCl

Volume of "Stock Solution" = 10.0 mL

Molarity of Final Solution = ?

Explanation - (1.0 M NaCl x 0.01 L) / 0.07621 L = M2

M1V1/V2 = M2

Calculation - Find molarity of NaCl

Part C - Preparation of 100.00 mL solution with a specified mass of alum, KAI(SO4)2.12H2O. Student will determine the molarity of a particular ion (K+, Al+3, SO4-2) in the resulting solution (K+, Al+2, SO4-2)

Answer = 0.078 M Na Cl

Explanation - 1.84 g Alum (1 mole/474.39 g Alum) = 3.8787x10^-3 mole

3.8787x10^-3 x 2 = 7.7573 x 10^-3 mole

7.7573 x 10^-3 / 0.1 L = 0.078 M NaCL

Calculation - Molarity of ion in final solution

Part D - Preparation of a sucrose solution at a specified % (m/m0 sucrose and a specified total mass of solution

Assigned Mass of Solution = 35.3 g

Assigned % by Mass Mass of sucrose = ? Mass of Water = ?

Answers = 25.5 g Sucrose and 9.7 g H20

Explanation - Suroces = 35.2 g x 0.725 (m/m)

Water = 35.2 g - 25.5 g

Calculation - Masses of sucrose and water required (dry lab)

Part E - Preparation of a solution with specified molality of sucrose using a specified mass of water

Assigned Molality = 0.339 g

Assigned Mass of Water = 52.2 g

Mass of Sucrose = ?

Mass of Prepared Solution = ?

Calculation - Masses of sucrose required in solution (dry lab)

Part F - Prepare a standard solution of KIO3 for Experiment 18

Volume of Solution =

Mass of KIO3 = Molarity of KIO3 = Answer = Explanation -

Calculation - Molarity of prepared solution

(Important to use deionized water, since solution will be used as a standard)

### Sets with similar terms

Chem 221 Podcast Lecture Notes (chapter 2, 3.…

50 terms ChelseaOlson30 Solution Problems 22 terms Bening-MCHS

AP Chemistry Test 2 - CONCENTRATION AND RXNS

31 terms taylorbuote

Chemistry Ch 12 Study Guide

70 terms kikaydragon

### Sets found in the same folder

Solutions Test 85 terms Jane_Schlesinger Chem 2 Finals 24 terms katie_pitchford3 NATE Heat Pump Exam 50 terms Images dddpope PLUS Matter and Energy 30 terms cohen0525

### Other sets by this creator

DCT Practice 19 terms m_sanchez0512

Foundations of Decentralized Clinical Trials…

25 terms m_sanchez0512

Intro to Decentralized Clinical Trials (DCTs)

75 terms m_sanchez0512

fuente : quizlet.com

## Calculate the number of moles of sulfuric acid that is contained in 0.250 mL of 0.500 M sulfuric acid

Calculate the number of moles of sulfuric acid that is contained in 0.250 mL of 0.500 M sulfuric acid solution… Get the answers you need, now! 01/30/2020 Chemistry High School

Calculate the number of moles of sulfuric acid that is contained in 0.250 mL of 0.500 M sulfuric acid solution.

Ace 10.3K answers 37.7M people helped

The number of moles of the sulfuric acid in the given new volume of the acid is 1.25 x 10⁻⁴ mole.

The given parameters;

volume of the sulfuric acid, V₁ = 0.25 ml

concentration of the acid, C₁ = 0.5 M

The concentration of the sulfuric acid is calculated as follows; Thus, the number of moles of the sulfuric acid in the given new volume of the acid is 1.25 x 10⁻⁴ mole.

Firstly, we need to know the meaning of the 0.5M. This simply means 0.5 moles are present in 1L or 1dm^3. 0.5M could also be written as 0.5moles/L.

Now, we know that there are 0.5 moles in 1L, we now need to calculate the number of moles in 0.5ml.

It is necessary to know that 1000ml = 1L.

Now if 0.5 moles = 1000ml

xmoles = 0.25ml

X = (0.25 * 0.5)/1000 = 0.000125 moles

Need a bit more clarification? Get a high-quality answer with step-by-step explanations from a professional in just minutes instead!

### Still have questions?

Snap questions with the app

Get help from the community

Find expert explanations for textbooks

View instant step-by-step math solutions

## You might be interested in

How many moles of sulfuric acid are contained in 100.0 mL of a 0.250 M sulfuric acid solution?

What volume of a 0.160 m solution of koh must be added to 450.0 ml of the acidic solution to completely neutralize all of the acid?

An acid solution is 0.100 M in HCl and 0.210 M in H2SO4. What volume of a 0.150 M solution of KOH must be added to 500.0 mL of the acidic solution to completely neutralize all of the acid?

if a 25.0 ml sample of sulfuric acid is titrated with 50.0 ml of 0.200 M potassium hydroxide to a phenolphthalein endpoint, what is the molarity of the acid? A) 0.150 M B) 0.100 M C) 0.200 M D) 0.300 M E) 0.400 M

### New questions in Chemistry

True or false? A polymer is a substance made of macromolecules.

How is a covalent structure formed

Why is a low-pressure system generally associated with cloudy and rainy weather?.

Which atom’s ionization energy is greater than that of phosphorus (P)? A) Ba B) K C) As D) Cl

Chemical compound responsible for ocean acidification in the open oceans of the world.

Previous Next

fuente : brainly.com

¿Quieres ver la respuesta o más?
Santiago 5 month ago

Chicos, ¿alguien sabe la respuesta?