The relative volumes of chloroform and water that should be used is 9:10
Concentration of solution in chloroform = 
( moles of chloroform )
Concentration of solution in water = 
( moles of water )
Dissociation constant at 
; 
Concentration of solution in chloroform / Concentration of solution in water
Meaning;
Since mole is present in chloroform and mole is present in water, Total mole of Caffeine present is 
Now, we substitute our given values into the equation
Therefore, the relative volumes of chloroform and water that should be used is 9:10
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The correct answer is A, since it is the only sensible answer.
<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>
Radioactive decay => C = Co { e ^ (- kt) |
Data:
Co = 2.00 mg
C = 0.25 mg
t = 4 hr 39 min
Time conversion: 4 hr 39 min = 4.65 hr
1) Replace the data in the equation to find k
C = Co { e ^ (-kt) } => C / Co = e ^ (-kt) => -kt = ln { C / Co} => kt = ln {Co / C}
=> k = ln {Co / C} / t = ln {2.00mg / 0.25mg} / 4.65 hr = 0.44719
2) Use C / Co = 1/2 to find the hallf-life
C / Co = e ^ (-kt) => -kt = ln (C / Co)
=> -kt = ln (1/2) => kt = ln(2) => t = ln (2) / k
t = ln(2) / 0.44719 = 1.55 hr.
Answer: 1.55 hr
When 0.514 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.8 C to 29.4 C. Find ⌂E rxn for the combustion of biphenyl in kJ/mol biphenyl. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.86 kJ/ C.
<span>The answer is - 6.30 * 10^3 kJ/mol
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