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How many mL of 10% w/v NaCl solution is needed to prepare 500 mL of a 0.8 % w/v NaCl solution?
500*0,008= 4 ml of 100% NaCl4/0,1= 40 ml of 10% NaCl
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How many grams of NaCl is required prepare 200 mL of a solution of 1 M NaCl (Molar of mass of NaCl=58.44g/mol)?
200 ml of 1 M Nacl have Moles of NaCl = 200 × 1 / 1000 = 0.2 mole Mass of NaCl = moles of NaCl × Molar mass .Mass...
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Calculate the molarity of the NaCl NaCl solution when 14.6 g g of NaCl NaCl is dissolved in 100 mL mL of water.
Molar mass of NaCl= 58.4414.6/58.44=0.25mol0.25/0.1= 2.5mol/L
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an NaCl solution has a concentration of 132 ppm. What volume of solution will provide a mass of 0.024 g of NaCl
Volume of solution =375mL
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A sample of gas expands from 2.00L to 3.00L against a constant pressure of 1.0atm. Calculate the work done in joules. (1L.atm= 101.3J)
101.3 Jouls
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How many grams of NaCl should be weighed to prepare 1.00 L of 601 ppm solution of Na<sup>+</sup>?
1L = 1000gNaCl -> Na+ + Cl-Mole ratio is 1:1ppm = ( m solute/m water) × 106M solute = (ppm × m water)/106= (601 × 1000)/106= 0.601g= 0.601g/23.0g=0.02613 moles Na+Now...
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Calculate the maximum mass of NaIO3 that can be dissolved in 3.00L of 0.200M AgNO3 solution without forming a precipitate
Answer on Question #97473 – Chemistry – OtherTask:Calculate the maximum mass of NaIO3 that can be dissolved in 3.00L of 0.200M AgNO3 solution without forming a precipitate.Solution:NaIO3 + AgNO3 =...
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2 components combine to form 2 components. 0.1 mole of xyz weight 10 gm and 0.05 moles of xys weight 9 gm . calculate the atomic mass of xyz
100 g/molAtomic mass of xyz is equal to A (xyz) = m (xyz) / n (xyz) = 10/0.1 = 100 g/mo
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How many grams of NaOH must be added to a 1.00L aqueous solution containing 2.00g of NiCl2 to fully remove the nickel (II) cation from the solution in the form of a Ni(OH)2 precipitate?
NiCl2 + 2NaOH = Ni(OH)2↓ + 2NaCl;m(NiCl2)/M(NiCl2)=m(NaOH)/2*M(NaOH) ;m(NaOH)=m(NiCl2)*2*M(NaOH)/M(NiCl2) ;M(NaOH)=23+16+1=40;M(NiCl2)=59+2*35.5=130;m(NaOH)=2*2*40/130=1.23 gram;Answer: 1.23 grams of NaOH are needed to fully remove the nickel (II) cation from the solution.
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How many moles of air are in 1.00L at 14.0⁰C and 161KPa?
pV=nRTn=pV/RT=161 x 1 / 8.314 x 287=0.0674736118 moles.
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