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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 mole fraction using volume fraction of 0.8629 specific gravity of .4040 and MW solute of 59.00
0.8629×0.4040= 0.34860.3486/59= 0.006g=6/1000g
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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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An aqueous solution has 18% of solute. By chemical analysis, it was known that it has 10 mol% solute.
What is the molecular weight of the solute?
m = n*M, m - weight, n - quantity of substance, M - molecular weight mx /18 = mH2O / 82nx = nH2O / 9nx *Mx = nH2O * MH2O*18/82Mx...
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I want to achieve the 10mcg concentration per 60ml of juice
The powder spec is
2500mcg/g concentration
Molecular mass 384.65g/mol
In the question of many unknown quantities. To prepare a solution of a given concentration you need 0.004 g of powder spec.
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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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Calculate the weight of water in a 1.5 liters 95% nitric acid (HNO3) that has a specific gravity of 1.82. Express the concentration of solution in terms of Molarity, Normality and molality.
1) density = m/VMass of HNO3 = density x V = 1.82 x 1.5 = 2.73 kg% = (100 x m(solvent))/ (m(solvent) + m(solute))m(HNO3) = 95/100 x 2.73 = 2.59...
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Calculate the concentration of OH- ions when reacting with 100 cm3 of KOH solution, concentration 0.09 mol / dm3 with 120 cm3 of H2SO4 solution with concentration 0.1 mol / dm3!
The reaction given:2KOH + H2SO4 = K2SO4 + 2H2OAs 1 L = 1dm3 and 1 ml = 1 cm3, the number of moles of the reactants equals:n(KOH) = M × V = 0.09...
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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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