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You are to make 10.0 L of 1.20 M Potassium nitrate solution. What molarity would the potassium nitrate solution need to be if you were to use only 2.50 L of it?
Molarity needed for Potassium Nitrate Solution ;=4.8M
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X cm³ of 0.15 mol dm³ zinc nitrate solution Zn(NO3)2 is pipetted into a 500 cm³ volumetric flask to produce 500 cm³ of 0.018 mol dm³ zinc nitrate solution. Zn(NO3)2. Determine the value of X.
Molar mass of Zinc nitrate = 189.36 189.36 × 0.018 =3.41×10 = 34.1= 500× 0.15/34.1= 2.2cm3
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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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What volume of 0.53 mol per DM cube of Potassium Hydroxide would yield 8.2 gram of potassium hydroxide in one vaporization to dryness
CM = n/VV = n / CMn = m / MM (KOH) = 56.1 g/moln (KOH) = 8.2 / 56.1 = 0.15 molV (KOH) = 0.15 / 0.53 = 0.28...
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How many grams of solid will precipitate if 26.0 mL of 0.285 mol/L magnesium nitrate are combined with 32.0 mL of 0.265 mol/L potassium hydroxide?
Solution:magnesium nitrate - Mg(NO3)2potassium hydroxide - KOHBalanced chemical equation:Mg(NO3)2(aq) + 2KOH(aq) → Mg(OH)2(s) + 2KNO3(aq)According to the equation above: n(Mg(NO3)2) = n(KOH)/2 = n(Mg(OH)2)26.0 mL of a 0.285 M solution...
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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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What is the pressure inside a cylinder when the volume changes from 14.37 dm3 to 49.31 dm3 with an initial pressure of 9.021 x 102 kPa? The temperature of the cylinder remained constant at 22.06ºC.
Answer on Question #83633, Chemistry/ General ChemistryWhat is the pressure inside a cylinder when the volume changes from 14.37 dm3 to 49.31 dm3 with an initial pressure of 9.021 x...
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A 0.4230 g sample of impure sodium nitrate was heated, converting all the sodium nitrate to .1235 g of sodium nitrate and oxygen gas. Determine the percent of sodium nitrate in the original sample.
Answer:2HNO3 = 2HNO2 + O2m= 85 x 0.1235 / 69= 0.1521W%= 0.1521 / 0.4230 = 36%
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