Answer:
56.8cm³
Explanation:
Given parameters:
Measured volume = 54.5cm³
Percentage error = 4.25%
Unknown:
Actual volume of the cylinder = ?
Solution:
The percentage error shows the amount of error introduced into a measurement.
We need to find out this amount of error from the data given.
Error = Percentage error x measured volume
Error = x 54.5 = ±2.32cm³
Since the error introduced = 2.32cm³
Actual volume of cylinder = measured volume ± Error
Since the percentage error quoted as higher 4.25%;
Actual volume = 54.5cm³ + 2.32cm³ = 56.8cm³
<span>10.2 grams of ethyl butyrate synthesized.
The balanced equation for the reaction of butanoic acid (C4H8O2) with ethanol (C2H6O) to produce ethyl butyrate (C6H12O2) is:
C4H8O2 + C2H6O ==> C6H12O2 + H2O
So for each mole of C4H8O2 used, 1 mole of C6H12O2 will be produced. So let's calculate the reactant and product molar masses. Start by looking up the atomic weights of the involved elements:
Atomic weight carbon = 12.0107
Atomic weight hydrogen = 1.00794
Atomic weight oxygen = 15.999
Molar mass C4H8O2 = 4 * 12.0107 + 8 * 1.00794 + 2 * 15.999 = 88.10432 g/mol
Molar mass C6H12O2 = 6 * 12.0107 + 12 * 1.00794 + 2 * 15.999 = 116.15748 g/mol
Moles C4H8O2 = 7.75 g / 88.10432 g/mol = 0.087963905 mol
Mass C6H12O2 = 0.087963905 mol * 116.15748 g/mol = 10.21766549 g
Rounding to 3 significant figures gives 10.2 grams of ethyl butyrate synthesized.</span>
The complete balanced chemical reaction for this is:
C6H8O7 + 3 NaHCO3 ---> 3 H2O + 3 CO2 + Na3C6H5O7
First calculate for the number of moles of each reactant.
moles C6H8O7 = 1 g / (192.124 g / mol) = 5.2 * 10^-3 mol
moles NaHCO3 = 1 g / (84.01 g / mol) = 11.9 * 10^-3 mol
A. The ratio of the reactant from the chemical reaction
is 3NaHCO3:1C6H8O7, while the given chemicals are in the ratio of:
11.9 * 10^-3NaHCO3: 5.2 * 10^-3 C6H8O7 =
2.29NaHCO3:1C6H8O7
Therefore this means that there is less amount of NaHCO3
supplied than what is required therefore the limiting reactant is:
NaHCO3
B. We calculate based on the limiting reactant.
mass CO2 = 11.9 * 10^-3 mol NaHCO3 (3 mol CO2/1 mol
NaHCO3) (44.01 g/mol)
mass CO2 = 1.57 g
C. I believe what is asked here is the amount of excess
reactant which remains. The excess reactant is C6H8O7.
mass C6H8O7 left = [5.2 * 10^-3 mol – (11.9 * 10^-3 mol
NaHCO3 (1 mol C6H8O7/ 3 mol NaHCO3))] * (192.124 g / mol)
mass C6H8O7 left = 0.237 g
Answer:
Plants use carbon dioxide to make food.
Explanation: