How many grams of Na3PO4 required to prepare 8 N in 500 ml? O 218.6 gm. O 656 gm. O 323.6 gm.

Q: How many g of Al2(SO4)3 can you prepare from 2.5 g of H2SO4  ?

A: When H2SO4 react with Al(OH)3 it form Al2(SO4)3 and H22O as product.

Q: How many grams of H2SO4 would be required to prepare 750.0 mL of 0.0900 N H2SO4?

A: Normality of solution = mass of soluteEquivalent mass of solute ×volume of solution(in L)  0.0900…

Q: 2. What mass of NaCl is needed to prepare 30.0% by mass of 500g solution of NaCl? Solution: 30.0% by…

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Q: 4. For the following reaction, OH O OH salicylic acid C7H6O3 138.12 g/mol acetic anhydride C4H603…

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Q: How many liters of a 1.8 M K N O 3 solution can be prepared from 69.7 grams of K N O 3?  Round your…

A: Molar mass of any compound can be expressed in g mol-1 and atomic mass units. Every molecule has its…

Q: How many grams of C2H5OH (Molar mass 46.07) are needed to prepare 3.75 x 102 mL of a 0.0325 M C2H5OH…

A: The concentration of a solution is related to the number of moles and the volume of the solution by…

Q: 1. How many grams of NiCl2-6H,O will be used to prepare a 0.0350 M, 500.0 mL of NiCl2 solution?

A: “Since you have asked multiple question, we will solve the first question for you. If you want any…

Q: A 0.119 M MgCO3 solution is made from 6.09 g of solute? What is the volume of the solution?

A: Given that,Mass of magnesium carbonate =6.09gMolar mass of magnesium carbonate=84.3139g/mole

Q: Wine is approximately 12% ethanol (CH3CH2OH) by volume. Ethanol has a molar mass of 46.06 g/moland a…

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Q: What mass of solid NaOH (97.0% NaOH by mass) is required to prepare 1.00 L of a 10.0% solution of…

A: Given,97.0% NaOH by massDensity= 1.109 g/mL

Q: How many moles of NaOH are present in 27.0 mL of 0.110 M NaOH?

A: Molarity is one of the concentration term which is used to define the concentration of a solute in a…

Q: How many moles of NaOH are present in 21.6 mL of 1.00 M NaOH

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Q: How many moles do you need to prepare 1.481 L of a 1.77 M NaOH solution? Round your answer to 2…

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Q: What is the molarity of NH4Cl in a 158.9mL aqueous solution containing 11.3g of NH4Cl? What is the…

A: We have to calculate Molarity in Question 1, and the mass percentage of ZnCl2 in Question 2.

Q: 30. The molarity of a solution containing 2.5 mol of acetic acid, HC2H3O2, in 400. mL of solution is…

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Q: 12.0 g NaHSO4 (MM:120.1 g/mol) is dissolved in water to make a 2.00 L solution. What is the molarity…

A: Molarity is defined as the number of moles of solute in 1 litre of solution.

Q: How many grams of ASA, C9H8O4, are present in 250.0 mL of 3.0 x 10-3 M C9H8O4? The molar mass of…

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Q: Consider a 3.00 M CH3OH solution. What volume of this solution, in ml, is needed to provide 0.202 g…

A: Given that : Molarity of CH3OH = 3.0 M Mass of CH3OH = 0.202 g Molar mass of CH3OH = 32.042 g/mol…

Q: Calculate the mass percent of solute in each solution. Calculate the mass percent of 2.47 g KCl2.47…

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Q: An aqueous solution is to be prepared that will be 3.87% by mass ammonium nitrate. What mass of…

A: Given :  Mass percent of ammonium nitrate = 3.87% Mass of solution = 1.23 kg To find  Mass = g…

Q: Convert 25.00 mL of 0.1010 M NaOH to moles NAOH:

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Q: A solution of H2SO4 contains 65% w/w of the acid, and has a density of 1.55 g/mL. How many molecules…

A: Answer:- This question is answered by using the simple concept of calculation of number of molecules…

Q: How many grams of solute are present in 145 mL of 0.590 M KBr? mass of solute:             g

A: Number of moles of solute = Molarity x volume in L Mass in grams = Number of moles x Molar mass

Q: What mass of C6H12O6 (glucose) is needed to prepare 450. mL of a 0.650 M solution of glucose in…

A: As per given information in question Volume = 450 mLMolarity = 0.650 M = 0.650 mol/L

Q: 1-Calculate the mass of compound required to prepare 100.00 mL of a 1.40x102 M CuCl2 ?.

A: Molecular weight of CuCl2 = 134.45 g mol-1 Formula Used 1. No of moles = volume × concentration 2.…

Q: . Calculate the number of moles in these quantities: (a) 25.0 g KNO3 (c) 5.4 x 102 g (NH4)2C2O4 (b)…

A: We have given that  1. Calculate the number of moles in these quantities: (a) 25.0 g KNO3 (c) 5.4 x…

Q: What mass of H2SO4 would be required to prepare 750. mL of a 0.150 M H2SO4 solution?

A: Given, Molarity of H2SO4 solution = 0.150 M = 0.150 mol/L Volume of H2SO4 solution = 750. mL = 0.750…

Q: How many grams of water must be added to 15.0 g NaCl to make a solution that is 10.0%(m/m)?

A: NOTE : Since you've asked multiple questions,we'll solve first question for you. If you want any…

Q: 5. What volume of beer that is 6.0 % by volume alcohol contains 200. ml CH3CH,OH (ethyl alcohol)?

A: Given:Volume/Volume % = 6.0 %Volume of ethyl alcohol = 200.0 mL

Q: 2. How much 12.0 M HNO3 solution and how much water should be mixed together to prepare 2.25L of…

A: The dilution equation is given below.M1* V1 = M2* V2Where, M1 is the original concentration, M2 is…

Q: 1.) What is the molarity of a solution containing 15.6 g of KCl in 274 mL of KCl solution? 2.)…

A: Molarity ( M ) :-  The number of moles of solute present in one litre of solution is called molarity…

Q: Limestone is typically 50.0% (w/w) CaCO3. How many grams of CaCO3 are present in a limestone sample…

A: Here we are required to find the mass of calcium carbonate present in 1.2 kg

Q: How many grams of KNO3 will be required to prepare 800 mL of 1.40 M KNO3?

A: Molarity = number of moles of solute/volume of solution (in L)  so, moles = molarity × volume (in L)…

Q: What volume (in L) of 7.16 M NAOH solution must be diluted to prepare 2.8 L of 0.500 M NaOH…

A: The dilution formula is,  (M1)(V1) = (M2)(V2) Hence, (V1) = (M2/ M1) (V2) Where V1 is volume of…

Mole Concept

Mole concept is a method to determine the amount of any substance that consists of some elemental particles using a grouping unit called ‘mole’. The mole concept enables one to handle the large amount of small elementary particles. The mole concept can be used in many calculations related to mole calculation, molar amount, mole ratio, and so on, which are significant in chemistry.