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What is the mass fraction and volume fraction of an aqueous solution having 0.7304 Mol Fraction, 0.6268 Specific Gravity of solute and 42.00 Molar Weight of solute?
Molar mass of solvent = 42/ 0.6268= 66.8789Total mass= 66.8789 + 42= 108.8789Mass fraction= 42/108.8789= 0.3857Volume fraction = 0.3857/ 0.7304=0.528
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In concentration of Aqueous solutions find mass fraction and volume Fraction using the mol fraction of 0.7304, Specific Gravity of solute of 0.6268 and molecular weight of solute of 42.00
=35.3molal
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<ol><li>10 mol of Zn react with 10 mol of HCl. Calculate the number of moles of H2 produced.</li></ol> (a) 5 mol (b) 10 mol (c) 20 mol (d) 2.5 mo
ANSWER SHOUD BE "a".EXPLANATION ......Zn reacts with HCl and form ZnCl2 and H2 .Balance Reaction is Zn + 2HCl ----- ZnCl2 + H2 .When 10 mol of Zn react with...
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Use the data to find another value for the enthalpy of buta-l , 3 —diene.
Bond energies Enthalpies of atomisation
C-H 413 KJ/mol C 715 KJ/mol
346 KJ/mol H 218 KJ/mol
610 KJ/mol
∆Hf = {Σ ∆H to turn elements into gaseous atoms -Σ ∆H to turn compound into gaseous atoms}4C(s) + 3H2(g) =CH2CHCHCH2(g)∆ Hf = {4x∆Hat [C] + 3 x E[H-H]) –...
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Which solution has the lowest pOH? 5.00x10-1 mol/L NH3 5.00x10-1 mol/L KOH 8.00x10-2 mol/L F- 8.00x10-2 mol/L Ca(OH)2 Chat started
First, we need to write the chemical equation. HCN --> H+ + CN-Next, make an ICE chart with the initial concentrations of [HCN] = 0.0620M and [H+] = [CN-] =...
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how will you prepare 50mills of 50 millimoles solution of potassium hydroxide and calcium hydroxide given that their molecular weight are 56.1g/mol and 74.0g/mol respectively
CM = n/Vn = m/M1) 0.05 M = n/0.05 Ln (KOH) = 0.05/0.05 = 1 molm (KOH) = n x M = 1 x 56.1 = 56.1 gTo prepare 50mills...
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From an initial mix of: 2 mol Cl2, 1 mol Br2, 0.5 mol I2, 0.5 mol Cl−, 1 mol Br−, 2 mol I−; the total quantity of I2 present after all possible reactions occur is
0.5 mol I2 and 0.5 mol Cl- will not react with anything; 2 mol I- will react with one mol of Cl2 producing one mol of I2: 2I- + Cl2...
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I have 2.65 L of an aqueous solution of NaBr. An excess of AgNO3 is added. 7.99 g of AgBr (s) are formed. What was the initial concentration of Br- (in mol / L) in this aqueous solution of NaBr?
0,016 mol / L
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50 ml of HCl solution (1 mol/L) were added to 53 g of AgNO3 solution (0.22 w/w%). How many grams of AgCl were produced? The molecular masses are 169.9 g/mol (AgNO3) and 143.3 g/mol (AgCl).
HCl + AgNO3----------- AgCl + HNO3To get our limiting reagent that will be used for the reaction:Using n=CV for HclC=1mol/L V=50/1000= 0.05Ln=1×0.05=0.05molesTherefore using a mole-mass relationship If 1mole of HCl...
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The emperical formula of a compound is C5H11. What is R the ration of the molecular weight to the emperical weight, if the molecular weight of the compound is 142 mg/mmol
Mr(C5H11) = 12 x 5 + 1 x 11 = 71 g/molR = 142 g/mol / 71 g/mol = 2C10H22
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