In the crystallization process the solid compound is dissolved in the solvent at elevated temperature and the crystallize product obtained by slow cooling of the solution. Here the solubility of acetanilide at 100°C is 1g per 20mL of water. Thus to dissolve 500mg of acetanilide at high temperature that is 100°C we need 10mL of water.
Now at 25°C after the re-crystallization there will be some amount of dissolve acetanilide. Which can be calculated as- 185mL of water is needed to dissolve 1g or 1000mg of acetanilide at 25°C. Thus in 10mL of water there will be 
gmg of acetanilide.
<span>C2H5
First, you need to figure out the relative ratios of moles of carbon and hydrogen. You do this by first looking up the atomic weight of carbon, hydrogen, and oxygen. Then you use those atomic weights to calculate the molar masses of H2O and CO2.
Carbon = 12.0107
Hydrogen = 1.00794
Oxygen = 15.999
Molar mass of H2O = 2 * 1.00794 + 15.999 = 18.01488
Molar mass of CO2 = 12.0107 + 2 * 15.999 = 44.0087
Now using the calculated molar masses, determine how many moles of each product was generated. You do this by dividing the given mass by the molar mass.
moles H2O = 11.5 g / 18.01488 g/mole = 0.638361 moles
moles CO2 = 22.4 g / 44.0087 g/mole = 0.50899 moles
The number of moles of carbon is the same as the number of moles of CO2 since there's just 1 carbon atom per CO2 molecule.
Since there's 2 hydrogen atoms per molecule of H2O, you need to multiply the number of moles of H2O by 2 to get the number of moles of hydrogen.
moles C = 0.50899
moles H = 0.638361 * 2 = 1.276722
We can double check our math by multiplying the calculated number of moles of carbon and hydrogen by their respective atomic weights and see if we get the original mass of the hydrocarbon.
total mass = 0.50899 * 12.0107 + 1.276722 * 1.00794 = 7.400185
7.400185 is more than close enough to 7.40 given rounding errors, so the double check worked.
Now to find the empirical formula we need to find a ratio of small integers that comes close to the ratio of moles of carbon and hydrogen.
0.50899 / 1.276722 = 0.398669
0.398669 is extremely close to 4/10, so let's reduce that ratio by dividing both top and bottom by 2 giving 2/5.
Since the number of moles of carbon was on top, that ratio implies that the empirical formula for this unknown hydrocarbon is
C2H5</span>
To answer the problem above first we need to find the difference of molar mass of NI3 from I2, 394.71 g/mol - 253.80 g/mol = 140.91 g/mol. Knowing the molar mass of the difference of NI3 from I2, in equation mass (g) / moles (mol) = molar mass, then we substitute. 3.58g / moles = 140.91 g/mol.
moles = 3.58 / 140.91 = 0.025 moles.
Answer:
Around 450 B.C.
Explanation:
The idea was forgotten until the 1800 when John Dalton re-introduced the atom.
Answer:
V HCNsln = 0.9176 L
Explanation:
V HCNsln = ?
∴ m HCN = 31 g
∴ <em>C</em> HCNsln = 1.25 mol/L
∴ molar mass HCN = 27.0253 g/mol
⇒ V HCNsln = (31 g)*(mol/27.0253 g)*(L/1.25 mol) = 0.9176 Lsln
