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Calculate the mole fraction of H2SO4 in 8%(w/w) aqueous H2SO4 ( molar mass: 98 g/mol, 18g/mol respectively
8%(w/w) aqueous H2SO4 solution means that 100 g of solution contains 8 g of the acidMass of water: 100 – 8 = 92 (g)Moles of each: H2SO4: 8 g /...
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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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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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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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A solution contains a solute with molar mass 94.5 g/mol. The solution is
1.77 m and also 1.70 M. Calculate the density of the solution.
c - molarity = 1.70 M b - molality = 1.77 m M - molar mass = 94.5 g/mol d - density - g/mL c = n(solute) / V(solution) n(solute)...
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How many molecules of hydrogen gas are there in 8.32 g of Hepatanoic acid? Molar Mass of the Acid = 130.182 g/mol Molar Mass of Hydrogen = 1.008 g/mol
Solution:Heptanoic acid = C7H14O2Hydrogen gas = H2C7H14O2 → 7H2According to the scheme above: n(C7H14O2) = n(H2)/7Thus:(8.32 g C7H14O2) × (1 mol C7H14O2 / 130.182 g C7H14O2) × (7 mol H2...
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Calculate number of moles in
A)7.85gm Fe (molar mass =56 gm/mol)
B)4.68 mg of Si (molar mass =28.1gm/mol)
n=m/M A. n = 7.85g / 56g/mol = 0.14 mol B. n = 0.00468g / 28.1 g/mol = 0.00017
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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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The molar mass of boron atoms in a natural sample is 10.7g/mol. The sample is known to consist of 10B (molar mass 10.013 g/mol). What is the percentage abundance of 10B?
10.013 x 100% / 10.7 = 93.5794393 %
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An aqueous solution contains 0.87 g/L of an organic compound. The osmotic pressure of this solution is 83 mmHg at 29°C. What is the molar mass of the solute?
Posm=CM*R*T; [kPa]Posm=83 mmHg =11,1 kPaR=8.314 J/(mol*K)T=29°C=273+29=302 KCM=n/V=m/(Mr*V)If V=1, m=0.87 g, thenCM=0.87/MrPosm=0.87/(Mr)*R*T;Mr=0.87*8.314*302/11.1=196.8=197 g/molAnswer: the molar mass of the solute is 197 g/mol
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